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Provide tags and a correct Python 3 solution for this coding contest problem. Tokitsukaze is playing a game derivated from Japanese mahjong. In this game, she has three tiles in her hand. Each tile she owns is a suited tile, which means it has a suit (manzu, pinzu or souzu) and a number (a digit ranged from 1 to 9). In this problem, we use one digit and one lowercase letter, which is the first character of the suit, to represent a suited tile. All possible suited tiles are represented as 1m, 2m, …, 9m, 1p, 2p, …, 9p, 1s, 2s, …, 9s. In order to win the game, she must have at least one mentsu (described below) in her hand, so sometimes she should draw extra suited tiles. After drawing a tile, the number of her tiles increases by one. She can draw any tiles she wants, including those already in her hand. Do you know the minimum number of extra suited tiles she needs to draw so that she can win? Here are some useful definitions in this game: * A mentsu, also known as meld, is formed by a koutsu or a shuntsu; * A koutsu, also known as triplet, is made of three identical tiles, such as [1m, 1m, 1m], however, [1m, 1p, 1s] or [1m, 4m, 7m] is NOT a koutsu; * A shuntsu, also known as sequence, is made of three sequential numbered tiles in the same suit, such as [1m, 2m, 3m] and [5s, 7s, 6s], however, [9m, 1m, 2m] or [1m, 2p, 3s] is NOT a shuntsu. Some examples: * [2m, 3p, 2s, 4m, 1s, 2s, 4s] β€” it contains no koutsu or shuntsu, so it includes no mentsu; * [4s, 3m, 3p, 4s, 5p, 4s, 5p] β€” it contains a koutsu, [4s, 4s, 4s], but no shuntsu, so it includes a mentsu; * [5p, 5s, 9m, 4p, 1s, 7p, 7m, 6p] β€” it contains no koutsu but a shuntsu, [5p, 4p, 6p] or [5p, 7p, 6p], so it includes a mentsu. Note that the order of tiles is unnecessary and you can assume the number of each type of suited tiles she can draw is infinite. Input The only line contains three strings β€” the tiles in Tokitsukaze's hand. For each string, the first character is a digit ranged from 1 to 9 and the second character is m, p or s. Output Print a single integer β€” the minimum number of extra suited tiles she needs to draw. Examples Input 1s 2s 3s Output 0 Input 9m 9m 9m Output 0 Input 3p 9m 2p Output 1 Note In the first example, Tokitsukaze already has a shuntsu. In the second example, Tokitsukaze already has a koutsu. In the third example, Tokitsukaze can get a shuntsu by drawing one suited tile β€” 1p or 4p. The resulting tiles will be [3p, 9m, 2p, 1p] or [3p, 9m, 2p, 4p].
instruction
0
5,106
19
10,212
Tags: brute force, implementation Correct Solution: ``` l = input().split() suit = ['m', 'p', 's'] from collections import defaultdict cnt = defaultdict(lambda : 0) for i in range(3): cnt[l[i]] += 1 mini = 10 for i in suit: for j in range(1, 10): # shuntsu if j + 2 <= 9: cn = 0 for k in range(3): cn += int("{}{}".format(j+k, i) not in cnt) mini = min(mini, cn) # koutsu mini = min(mini, 3 - cnt["{}{}".format(j, i)]) print(mini) ```
output
1
5,106
19
10,213
Provide tags and a correct Python 3 solution for this coding contest problem. Tokitsukaze is playing a game derivated from Japanese mahjong. In this game, she has three tiles in her hand. Each tile she owns is a suited tile, which means it has a suit (manzu, pinzu or souzu) and a number (a digit ranged from 1 to 9). In this problem, we use one digit and one lowercase letter, which is the first character of the suit, to represent a suited tile. All possible suited tiles are represented as 1m, 2m, …, 9m, 1p, 2p, …, 9p, 1s, 2s, …, 9s. In order to win the game, she must have at least one mentsu (described below) in her hand, so sometimes she should draw extra suited tiles. After drawing a tile, the number of her tiles increases by one. She can draw any tiles she wants, including those already in her hand. Do you know the minimum number of extra suited tiles she needs to draw so that she can win? Here are some useful definitions in this game: * A mentsu, also known as meld, is formed by a koutsu or a shuntsu; * A koutsu, also known as triplet, is made of three identical tiles, such as [1m, 1m, 1m], however, [1m, 1p, 1s] or [1m, 4m, 7m] is NOT a koutsu; * A shuntsu, also known as sequence, is made of three sequential numbered tiles in the same suit, such as [1m, 2m, 3m] and [5s, 7s, 6s], however, [9m, 1m, 2m] or [1m, 2p, 3s] is NOT a shuntsu. Some examples: * [2m, 3p, 2s, 4m, 1s, 2s, 4s] β€” it contains no koutsu or shuntsu, so it includes no mentsu; * [4s, 3m, 3p, 4s, 5p, 4s, 5p] β€” it contains a koutsu, [4s, 4s, 4s], but no shuntsu, so it includes a mentsu; * [5p, 5s, 9m, 4p, 1s, 7p, 7m, 6p] β€” it contains no koutsu but a shuntsu, [5p, 4p, 6p] or [5p, 7p, 6p], so it includes a mentsu. Note that the order of tiles is unnecessary and you can assume the number of each type of suited tiles she can draw is infinite. Input The only line contains three strings β€” the tiles in Tokitsukaze's hand. For each string, the first character is a digit ranged from 1 to 9 and the second character is m, p or s. Output Print a single integer β€” the minimum number of extra suited tiles she needs to draw. Examples Input 1s 2s 3s Output 0 Input 9m 9m 9m Output 0 Input 3p 9m 2p Output 1 Note In the first example, Tokitsukaze already has a shuntsu. In the second example, Tokitsukaze already has a koutsu. In the third example, Tokitsukaze can get a shuntsu by drawing one suited tile β€” 1p or 4p. The resulting tiles will be [3p, 9m, 2p, 1p] or [3p, 9m, 2p, 4p].
instruction
0
5,107
19
10,214
Tags: brute force, implementation Correct Solution: ``` # your code goes here a,b,c=input().split() d={'m':[],'p':[],'s':[]} d[a[1]].append(int(a[0])) d[b[1]].append(int(b[0])) d[c[1]].append(int(c[0])) l=['m','p','s'] ans=2 for i in l: if d[i]==[]: continue for j in range(1,10): ans=min(ans,3-d[i].count(j)) if ans<0: ans=0 if ans==0: break for j in range(1,8): if j in d[i] and j+1 in d[i] and j+2 in d[i]: ans=0 if ans==0: break if j in d[i] and j+1 in d[i] and j+2 not in d[i]: ans=1 if j in d[i] and j+2 in d[i] and j+1 not in d[i]: ans=1 if j not in d[i] and j+1 in d[i] and j+2 in d[i]: ans=1 if ans==0: break print(ans) ```
output
1
5,107
19
10,215
Provide tags and a correct Python 3 solution for this coding contest problem. Tokitsukaze is playing a game derivated from Japanese mahjong. In this game, she has three tiles in her hand. Each tile she owns is a suited tile, which means it has a suit (manzu, pinzu or souzu) and a number (a digit ranged from 1 to 9). In this problem, we use one digit and one lowercase letter, which is the first character of the suit, to represent a suited tile. All possible suited tiles are represented as 1m, 2m, …, 9m, 1p, 2p, …, 9p, 1s, 2s, …, 9s. In order to win the game, she must have at least one mentsu (described below) in her hand, so sometimes she should draw extra suited tiles. After drawing a tile, the number of her tiles increases by one. She can draw any tiles she wants, including those already in her hand. Do you know the minimum number of extra suited tiles she needs to draw so that she can win? Here are some useful definitions in this game: * A mentsu, also known as meld, is formed by a koutsu or a shuntsu; * A koutsu, also known as triplet, is made of three identical tiles, such as [1m, 1m, 1m], however, [1m, 1p, 1s] or [1m, 4m, 7m] is NOT a koutsu; * A shuntsu, also known as sequence, is made of three sequential numbered tiles in the same suit, such as [1m, 2m, 3m] and [5s, 7s, 6s], however, [9m, 1m, 2m] or [1m, 2p, 3s] is NOT a shuntsu. Some examples: * [2m, 3p, 2s, 4m, 1s, 2s, 4s] β€” it contains no koutsu or shuntsu, so it includes no mentsu; * [4s, 3m, 3p, 4s, 5p, 4s, 5p] β€” it contains a koutsu, [4s, 4s, 4s], but no shuntsu, so it includes a mentsu; * [5p, 5s, 9m, 4p, 1s, 7p, 7m, 6p] β€” it contains no koutsu but a shuntsu, [5p, 4p, 6p] or [5p, 7p, 6p], so it includes a mentsu. Note that the order of tiles is unnecessary and you can assume the number of each type of suited tiles she can draw is infinite. Input The only line contains three strings β€” the tiles in Tokitsukaze's hand. For each string, the first character is a digit ranged from 1 to 9 and the second character is m, p or s. Output Print a single integer β€” the minimum number of extra suited tiles she needs to draw. Examples Input 1s 2s 3s Output 0 Input 9m 9m 9m Output 0 Input 3p 9m 2p Output 1 Note In the first example, Tokitsukaze already has a shuntsu. In the second example, Tokitsukaze already has a koutsu. In the third example, Tokitsukaze can get a shuntsu by drawing one suited tile β€” 1p or 4p. The resulting tiles will be [3p, 9m, 2p, 1p] or [3p, 9m, 2p, 4p].
instruction
0
5,108
19
10,216
Tags: brute force, implementation Correct Solution: ``` s1,s2,s3 = input().split() A = [] if (s1[1]==s2[1])and(s2[1]==s3[1]): if (s1[0]==s2[0])and(s2[0]==s3[0]): print(0) exit() A.append(int(s1[0])) A.append(int(s2[0])) A.append(int(s3[0])) A.sort() if (A[0]==A[1]-1) and(A[0]==A[2]-2): print(0) exit() if (s1[1]==s2[1]): if (s1[0]==s2[0]): print(1) exit() if (int(s1[0])==(int(s2[0])+1))or(int(s1[0])==(int(s2[0])-1)): print(1) exit() if (int(s1[0])==(int(s2[0])+2))or(int(s1[0])==(int(s2[0])-2)): print(1) exit() if (s1[1] == s3[1]): if (s1[0]==s3[0]): print(1) exit() if (int(s1[0])==(int(s3[0])+1))or(int(s1[0])==(int(s3[0])-1)): print(1) exit() if (int(s1[0])==(int(s3[0])+2))or(int(s1[0])==(int(s3[0])-2)): print(1) exit() if (s2[1]==s3[1]): if (s2[0]==s3[0]): print(1) exit() if (int(s2[0])==(int(s3[0])+1))or(int(s2[0])==(int(s3[0])-1)): print(1) exit() if (int(s2[0])==(int(s3[0])+2))or(int(s2[0])==(int(s3[0])-2)): print(1) exit() print(2) ```
output
1
5,108
19
10,217
Provide tags and a correct Python 3 solution for this coding contest problem. Tokitsukaze is playing a game derivated from Japanese mahjong. In this game, she has three tiles in her hand. Each tile she owns is a suited tile, which means it has a suit (manzu, pinzu or souzu) and a number (a digit ranged from 1 to 9). In this problem, we use one digit and one lowercase letter, which is the first character of the suit, to represent a suited tile. All possible suited tiles are represented as 1m, 2m, …, 9m, 1p, 2p, …, 9p, 1s, 2s, …, 9s. In order to win the game, she must have at least one mentsu (described below) in her hand, so sometimes she should draw extra suited tiles. After drawing a tile, the number of her tiles increases by one. She can draw any tiles she wants, including those already in her hand. Do you know the minimum number of extra suited tiles she needs to draw so that she can win? Here are some useful definitions in this game: * A mentsu, also known as meld, is formed by a koutsu or a shuntsu; * A koutsu, also known as triplet, is made of three identical tiles, such as [1m, 1m, 1m], however, [1m, 1p, 1s] or [1m, 4m, 7m] is NOT a koutsu; * A shuntsu, also known as sequence, is made of three sequential numbered tiles in the same suit, such as [1m, 2m, 3m] and [5s, 7s, 6s], however, [9m, 1m, 2m] or [1m, 2p, 3s] is NOT a shuntsu. Some examples: * [2m, 3p, 2s, 4m, 1s, 2s, 4s] β€” it contains no koutsu or shuntsu, so it includes no mentsu; * [4s, 3m, 3p, 4s, 5p, 4s, 5p] β€” it contains a koutsu, [4s, 4s, 4s], but no shuntsu, so it includes a mentsu; * [5p, 5s, 9m, 4p, 1s, 7p, 7m, 6p] β€” it contains no koutsu but a shuntsu, [5p, 4p, 6p] or [5p, 7p, 6p], so it includes a mentsu. Note that the order of tiles is unnecessary and you can assume the number of each type of suited tiles she can draw is infinite. Input The only line contains three strings β€” the tiles in Tokitsukaze's hand. For each string, the first character is a digit ranged from 1 to 9 and the second character is m, p or s. Output Print a single integer β€” the minimum number of extra suited tiles she needs to draw. Examples Input 1s 2s 3s Output 0 Input 9m 9m 9m Output 0 Input 3p 9m 2p Output 1 Note In the first example, Tokitsukaze already has a shuntsu. In the second example, Tokitsukaze already has a koutsu. In the third example, Tokitsukaze can get a shuntsu by drawing one suited tile β€” 1p or 4p. The resulting tiles will be [3p, 9m, 2p, 1p] or [3p, 9m, 2p, 4p].
instruction
0
5,109
19
10,218
Tags: brute force, implementation Correct Solution: ``` import collections a=input().split() a.sort() if a[0]==a[1] and a[1]==a[2]: print(0) elif a[0][1]==a[1][1] and a[1][1]==a[2][1] and abs(int(a[1][0])-int(a[0][0]))==1 and abs(int(a[2][0])-int(a[1][0]))==1: print("0") elif a[0]==a[1] or a[1]==a[2] or a[2]==a[0]: print("1") elif a[0][1]==a[1][1] and (abs(int(a[1][0])-int(a[0][0]))==1 or abs(int(a[1][0])-int(a[0][0]))==2) : print("1") elif a[1][1]==a[2][1] and (abs(int(a[2][0])-int(a[1][0]))==1 or abs(int(a[2][0])-int(a[1][0]))==2): print("1") elif a[0][1]==a[2][1] and (abs(int(a[2][0])-int(a[0][0]))==1 or abs(int(a[2][0])-int(a[0][0]))==2): print("1") else: print("2") ```
output
1
5,109
19
10,219
Provide tags and a correct Python 3 solution for this coding contest problem. Tokitsukaze is playing a game derivated from Japanese mahjong. In this game, she has three tiles in her hand. Each tile she owns is a suited tile, which means it has a suit (manzu, pinzu or souzu) and a number (a digit ranged from 1 to 9). In this problem, we use one digit and one lowercase letter, which is the first character of the suit, to represent a suited tile. All possible suited tiles are represented as 1m, 2m, …, 9m, 1p, 2p, …, 9p, 1s, 2s, …, 9s. In order to win the game, she must have at least one mentsu (described below) in her hand, so sometimes she should draw extra suited tiles. After drawing a tile, the number of her tiles increases by one. She can draw any tiles she wants, including those already in her hand. Do you know the minimum number of extra suited tiles she needs to draw so that she can win? Here are some useful definitions in this game: * A mentsu, also known as meld, is formed by a koutsu or a shuntsu; * A koutsu, also known as triplet, is made of three identical tiles, such as [1m, 1m, 1m], however, [1m, 1p, 1s] or [1m, 4m, 7m] is NOT a koutsu; * A shuntsu, also known as sequence, is made of three sequential numbered tiles in the same suit, such as [1m, 2m, 3m] and [5s, 7s, 6s], however, [9m, 1m, 2m] or [1m, 2p, 3s] is NOT a shuntsu. Some examples: * [2m, 3p, 2s, 4m, 1s, 2s, 4s] β€” it contains no koutsu or shuntsu, so it includes no mentsu; * [4s, 3m, 3p, 4s, 5p, 4s, 5p] β€” it contains a koutsu, [4s, 4s, 4s], but no shuntsu, so it includes a mentsu; * [5p, 5s, 9m, 4p, 1s, 7p, 7m, 6p] β€” it contains no koutsu but a shuntsu, [5p, 4p, 6p] or [5p, 7p, 6p], so it includes a mentsu. Note that the order of tiles is unnecessary and you can assume the number of each type of suited tiles she can draw is infinite. Input The only line contains three strings β€” the tiles in Tokitsukaze's hand. For each string, the first character is a digit ranged from 1 to 9 and the second character is m, p or s. Output Print a single integer β€” the minimum number of extra suited tiles she needs to draw. Examples Input 1s 2s 3s Output 0 Input 9m 9m 9m Output 0 Input 3p 9m 2p Output 1 Note In the first example, Tokitsukaze already has a shuntsu. In the second example, Tokitsukaze already has a koutsu. In the third example, Tokitsukaze can get a shuntsu by drawing one suited tile β€” 1p or 4p. The resulting tiles will be [3p, 9m, 2p, 1p] or [3p, 9m, 2p, 4p].
instruction
0
5,110
19
10,220
Tags: brute force, implementation Correct Solution: ``` arr=list(map(str,input().split())) num1=int(arr[0][0]) num2=int(arr[1][0]) num3=int(arr[2][0]) str1=arr[0][1] str2=arr[1][1] str3=arr[2][1] if str1==str2 and str2==str3: if num1==num2 and num2==num3: print(0) else: arr2=[num1,num2,num3] arr2=sorted(arr2) if arr2[0]==arr2[1]-1 and arr2[1]==arr2[2]-1: print(0) else: if 0<=abs(num1-num2)<=2: print(1) elif 0<=abs(num1-num3)<=2: print(1) elif 0<=abs(num2-num3)<=2: print(1) else: print(2) elif str1==str2: if abs(num1-num2)==0 or abs(num1-num2)==1 or abs(num1-num2)==2: print(1) else: print(2) elif str1==str3: if abs(num1-num3)==0 or abs(num1-num3)==1 or abs(num1-num3)==2: print(1) else: print(2) elif str2==str3: if abs(num2-num3)==0 or abs(num2-num3)==1 or abs(num2-num3)==2: print(1) else: print(2) else: print(2) ```
output
1
5,110
19
10,221
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Tokitsukaze is playing a game derivated from Japanese mahjong. In this game, she has three tiles in her hand. Each tile she owns is a suited tile, which means it has a suit (manzu, pinzu or souzu) and a number (a digit ranged from 1 to 9). In this problem, we use one digit and one lowercase letter, which is the first character of the suit, to represent a suited tile. All possible suited tiles are represented as 1m, 2m, …, 9m, 1p, 2p, …, 9p, 1s, 2s, …, 9s. In order to win the game, she must have at least one mentsu (described below) in her hand, so sometimes she should draw extra suited tiles. After drawing a tile, the number of her tiles increases by one. She can draw any tiles she wants, including those already in her hand. Do you know the minimum number of extra suited tiles she needs to draw so that she can win? Here are some useful definitions in this game: * A mentsu, also known as meld, is formed by a koutsu or a shuntsu; * A koutsu, also known as triplet, is made of three identical tiles, such as [1m, 1m, 1m], however, [1m, 1p, 1s] or [1m, 4m, 7m] is NOT a koutsu; * A shuntsu, also known as sequence, is made of three sequential numbered tiles in the same suit, such as [1m, 2m, 3m] and [5s, 7s, 6s], however, [9m, 1m, 2m] or [1m, 2p, 3s] is NOT a shuntsu. Some examples: * [2m, 3p, 2s, 4m, 1s, 2s, 4s] β€” it contains no koutsu or shuntsu, so it includes no mentsu; * [4s, 3m, 3p, 4s, 5p, 4s, 5p] β€” it contains a koutsu, [4s, 4s, 4s], but no shuntsu, so it includes a mentsu; * [5p, 5s, 9m, 4p, 1s, 7p, 7m, 6p] β€” it contains no koutsu but a shuntsu, [5p, 4p, 6p] or [5p, 7p, 6p], so it includes a mentsu. Note that the order of tiles is unnecessary and you can assume the number of each type of suited tiles she can draw is infinite. Input The only line contains three strings β€” the tiles in Tokitsukaze's hand. For each string, the first character is a digit ranged from 1 to 9 and the second character is m, p or s. Output Print a single integer β€” the minimum number of extra suited tiles she needs to draw. Examples Input 1s 2s 3s Output 0 Input 9m 9m 9m Output 0 Input 3p 9m 2p Output 1 Note In the first example, Tokitsukaze already has a shuntsu. In the second example, Tokitsukaze already has a koutsu. In the third example, Tokitsukaze can get a shuntsu by drawing one suited tile β€” 1p or 4p. The resulting tiles will be [3p, 9m, 2p, 1p] or [3p, 9m, 2p, 4p]. Submitted Solution: ``` m={"m":[0]*9, "s":[0]*9, "p":[0]*9} for s in input().split(): m[s[1]][int(s[0])-1]+=1 ans = 2 for c in "smp": l = m[c] if(max(l)>=2): ans = min(ans, 3-max(l)) else: for i in range(7): sm = sum(l[i:i+3]) ans = min(ans, 3-sm) print(ans) ```
instruction
0
5,111
19
10,222
Yes
output
1
5,111
19
10,223
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Tokitsukaze is playing a game derivated from Japanese mahjong. In this game, she has three tiles in her hand. Each tile she owns is a suited tile, which means it has a suit (manzu, pinzu or souzu) and a number (a digit ranged from 1 to 9). In this problem, we use one digit and one lowercase letter, which is the first character of the suit, to represent a suited tile. All possible suited tiles are represented as 1m, 2m, …, 9m, 1p, 2p, …, 9p, 1s, 2s, …, 9s. In order to win the game, she must have at least one mentsu (described below) in her hand, so sometimes she should draw extra suited tiles. After drawing a tile, the number of her tiles increases by one. She can draw any tiles she wants, including those already in her hand. Do you know the minimum number of extra suited tiles she needs to draw so that she can win? Here are some useful definitions in this game: * A mentsu, also known as meld, is formed by a koutsu or a shuntsu; * A koutsu, also known as triplet, is made of three identical tiles, such as [1m, 1m, 1m], however, [1m, 1p, 1s] or [1m, 4m, 7m] is NOT a koutsu; * A shuntsu, also known as sequence, is made of three sequential numbered tiles in the same suit, such as [1m, 2m, 3m] and [5s, 7s, 6s], however, [9m, 1m, 2m] or [1m, 2p, 3s] is NOT a shuntsu. Some examples: * [2m, 3p, 2s, 4m, 1s, 2s, 4s] β€” it contains no koutsu or shuntsu, so it includes no mentsu; * [4s, 3m, 3p, 4s, 5p, 4s, 5p] β€” it contains a koutsu, [4s, 4s, 4s], but no shuntsu, so it includes a mentsu; * [5p, 5s, 9m, 4p, 1s, 7p, 7m, 6p] β€” it contains no koutsu but a shuntsu, [5p, 4p, 6p] or [5p, 7p, 6p], so it includes a mentsu. Note that the order of tiles is unnecessary and you can assume the number of each type of suited tiles she can draw is infinite. Input The only line contains three strings β€” the tiles in Tokitsukaze's hand. For each string, the first character is a digit ranged from 1 to 9 and the second character is m, p or s. Output Print a single integer β€” the minimum number of extra suited tiles she needs to draw. Examples Input 1s 2s 3s Output 0 Input 9m 9m 9m Output 0 Input 3p 9m 2p Output 1 Note In the first example, Tokitsukaze already has a shuntsu. In the second example, Tokitsukaze already has a koutsu. In the third example, Tokitsukaze can get a shuntsu by drawing one suited tile β€” 1p or 4p. The resulting tiles will be [3p, 9m, 2p, 1p] or [3p, 9m, 2p, 4p]. Submitted Solution: ``` import bisect import collections import copy import functools import heapq import itertools import math import random import re import sys import time import string from typing import * sys.setrecursionlimit(99999) arr = list(input().split()) arr.sort() ans = 3 cs = collections.Counter(arr) def check(ap): cp = collections.Counter(ap) c = 0 for k, v in cs.items(): c += min(v, cp[k]) return 3 - c for i in range(1, 10): ans = min(ans, check([str(i) + 's', str(i) + 's', str(i) + 's'])) ans = min(ans, check([str(i) + 'p', str(i) + 'p', str(i) + 'p'])) ans = min(ans, check([str(i) + 'm', str(i) + 'm', str(i) + 'm'])) if i <= 7: ans = min(ans, check([str(i) + 's', str(i + 1) + 's', str(i + 2) + 's'])) ans = min(ans, check([str(i) + 'p', str(i + 1) + 'p', str(i + 2) + 'p'])) ans = min(ans, check([str(i) + 'm', str(i + 1) + 'm', str(i + 2) + 'm'])) print(ans) ```
instruction
0
5,112
19
10,224
Yes
output
1
5,112
19
10,225
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Tokitsukaze is playing a game derivated from Japanese mahjong. In this game, she has three tiles in her hand. Each tile she owns is a suited tile, which means it has a suit (manzu, pinzu or souzu) and a number (a digit ranged from 1 to 9). In this problem, we use one digit and one lowercase letter, which is the first character of the suit, to represent a suited tile. All possible suited tiles are represented as 1m, 2m, …, 9m, 1p, 2p, …, 9p, 1s, 2s, …, 9s. In order to win the game, she must have at least one mentsu (described below) in her hand, so sometimes she should draw extra suited tiles. After drawing a tile, the number of her tiles increases by one. She can draw any tiles she wants, including those already in her hand. Do you know the minimum number of extra suited tiles she needs to draw so that she can win? Here are some useful definitions in this game: * A mentsu, also known as meld, is formed by a koutsu or a shuntsu; * A koutsu, also known as triplet, is made of three identical tiles, such as [1m, 1m, 1m], however, [1m, 1p, 1s] or [1m, 4m, 7m] is NOT a koutsu; * A shuntsu, also known as sequence, is made of three sequential numbered tiles in the same suit, such as [1m, 2m, 3m] and [5s, 7s, 6s], however, [9m, 1m, 2m] or [1m, 2p, 3s] is NOT a shuntsu. Some examples: * [2m, 3p, 2s, 4m, 1s, 2s, 4s] β€” it contains no koutsu or shuntsu, so it includes no mentsu; * [4s, 3m, 3p, 4s, 5p, 4s, 5p] β€” it contains a koutsu, [4s, 4s, 4s], but no shuntsu, so it includes a mentsu; * [5p, 5s, 9m, 4p, 1s, 7p, 7m, 6p] β€” it contains no koutsu but a shuntsu, [5p, 4p, 6p] or [5p, 7p, 6p], so it includes a mentsu. Note that the order of tiles is unnecessary and you can assume the number of each type of suited tiles she can draw is infinite. Input The only line contains three strings β€” the tiles in Tokitsukaze's hand. For each string, the first character is a digit ranged from 1 to 9 and the second character is m, p or s. Output Print a single integer β€” the minimum number of extra suited tiles she needs to draw. Examples Input 1s 2s 3s Output 0 Input 9m 9m 9m Output 0 Input 3p 9m 2p Output 1 Note In the first example, Tokitsukaze already has a shuntsu. In the second example, Tokitsukaze already has a koutsu. In the third example, Tokitsukaze can get a shuntsu by drawing one suited tile β€” 1p or 4p. The resulting tiles will be [3p, 9m, 2p, 1p] or [3p, 9m, 2p, 4p]. Submitted Solution: ``` #include<bits/stdc++.h> #include<stdio.h> //per fare input output con scanf and printf #include<stdlib.h> //per fare qsort e bsearch #include<string.h> // per fare strcpy(sarrivo, spartenza) strcat(str, aggiungo) strcmp(a,b) che da 0 se sono uguali #include<math.h> #include<algorithm> #include<iostream> #include<queue> #include<stack> #include<vector> #include<map> #using namespace std; #define ld long double #define ll long long int #define vi vector <ll> #define pi pair <ll, ll> #define binary(v, el) binary_search((v).begin(), (v).end(), (el)) #define PB push_back #define MP make_pair #define F first #define S second x,y,z = list(map(str, input().strip().split())) def cazzu(a,b,c): if a==b==c: return 1 else: return 0 def shizu(a,b,c): x = int(a[0]) y = int(b[0]) z = int(c[0]) l = [x,y,z] l.sort() if a[1]==b[1]==c[1] and l[1]-l[0] == 1 and l[2]-l[1]==1: return 1 else: return 0 if cazzu(x,y,z) or shizu(x,y,z): print(0) else: flag = 0 for i in ['p','m','s']: for j in ['1','2','3','4','5','6','7','8','9']: w = j+i if cazzu(x,y,w) or shizu(x,y,w) or cazzu(x,w,z) or shizu(x,w,z) or cazzu(w,y,z) or shizu(w,y,z): flag = 1 if flag == 1: print(1) else: print(2) ```
instruction
0
5,113
19
10,226
Yes
output
1
5,113
19
10,227
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Tokitsukaze is playing a game derivated from Japanese mahjong. In this game, she has three tiles in her hand. Each tile she owns is a suited tile, which means it has a suit (manzu, pinzu or souzu) and a number (a digit ranged from 1 to 9). In this problem, we use one digit and one lowercase letter, which is the first character of the suit, to represent a suited tile. All possible suited tiles are represented as 1m, 2m, …, 9m, 1p, 2p, …, 9p, 1s, 2s, …, 9s. In order to win the game, she must have at least one mentsu (described below) in her hand, so sometimes she should draw extra suited tiles. After drawing a tile, the number of her tiles increases by one. She can draw any tiles she wants, including those already in her hand. Do you know the minimum number of extra suited tiles she needs to draw so that she can win? Here are some useful definitions in this game: * A mentsu, also known as meld, is formed by a koutsu or a shuntsu; * A koutsu, also known as triplet, is made of three identical tiles, such as [1m, 1m, 1m], however, [1m, 1p, 1s] or [1m, 4m, 7m] is NOT a koutsu; * A shuntsu, also known as sequence, is made of three sequential numbered tiles in the same suit, such as [1m, 2m, 3m] and [5s, 7s, 6s], however, [9m, 1m, 2m] or [1m, 2p, 3s] is NOT a shuntsu. Some examples: * [2m, 3p, 2s, 4m, 1s, 2s, 4s] β€” it contains no koutsu or shuntsu, so it includes no mentsu; * [4s, 3m, 3p, 4s, 5p, 4s, 5p] β€” it contains a koutsu, [4s, 4s, 4s], but no shuntsu, so it includes a mentsu; * [5p, 5s, 9m, 4p, 1s, 7p, 7m, 6p] β€” it contains no koutsu but a shuntsu, [5p, 4p, 6p] or [5p, 7p, 6p], so it includes a mentsu. Note that the order of tiles is unnecessary and you can assume the number of each type of suited tiles she can draw is infinite. Input The only line contains three strings β€” the tiles in Tokitsukaze's hand. For each string, the first character is a digit ranged from 1 to 9 and the second character is m, p or s. Output Print a single integer β€” the minimum number of extra suited tiles she needs to draw. Examples Input 1s 2s 3s Output 0 Input 9m 9m 9m Output 0 Input 3p 9m 2p Output 1 Note In the first example, Tokitsukaze already has a shuntsu. In the second example, Tokitsukaze already has a koutsu. In the third example, Tokitsukaze can get a shuntsu by drawing one suited tile β€” 1p or 4p. The resulting tiles will be [3p, 9m, 2p, 1p] or [3p, 9m, 2p, 4p]. Submitted Solution: ``` ''' Auther: ghoshashis545 Ashis Ghosh College: jalpaiguri Govt Enggineering College ''' from os import path import sys from heapq import heappush,heappop from functools import cmp_to_key as ctk from collections import deque,defaultdict as dd from bisect import bisect,bisect_left,bisect_right,insort,insort_left,insort_right from itertools import permutations from datetime import datetime from math import ceil,sqrt,log,gcd def ii():return int(input()) def si():return input().rstrip() def mi():return map(int,input().split()) def li():return list(mi()) abc='abcdefghijklmnopqrstuvwxyz' mod=1000000007 # mod=998244353 inf = float("inf") vow=['a','e','i','o','u'] dx,dy=[-1,1,0,0],[0,0,1,-1] def bo(i): return ord(i)-ord('a') file=1 def solve(): # for _ in range(ii()): s = list(map(str,input().split())) m = {} m['s'] = [] m['p'] = [] m['m'] = [] for i in s: m[i[1]].append(int(i[0])) if len(set(s))==1: print(0) elif(len(set(s))==2): print(1) elif(len(m['s'])==3): p = [m['s'][0],m['s'][1],m['s'][2]] p.sort() if p[1]-p[0]==1 and p[2]-p[1]==1: print(0) elif(p[1]-p[0]==2 or p[2]-p[1]==2 or p[1]-p[0]==1 or p[2]-p[1]==1): print(1) else: print(2) elif(len(m['p'])==3): p = [m['p'][0],m['p'][1],m['p'][2]] p.sort() if p[1]-p[0]==1 and p[2]-p[1]==1: print(0) elif(p[1]-p[0]==2 or p[2]-p[1]==2 or p[1]-p[0]==1 or p[2]-p[1]==1): print(1) else: print(2) elif(len(m['m'])==3): p = [m['m'][0],m['m'][1],m['m'][2]] p.sort() if p[1]-p[0]==1 and p[2]-p[1]==1: print(0) elif(p[1]-p[0]==2 or p[2]-p[1]==2 or p[1]-p[0]==1 or p[2]-p[1]==1): print(1) else: print(2) elif(len(m['s'])==2 and (abs(m['s'][0]-m['s'][1]) == 1 or abs(m['s'][0]-m['s'][1])==2)): print(1) elif(len(m['p'])==2 and (abs(m['p'][0]-m['p'][1]) == 1 or abs(m['p'][0]-m['p'][1])==2)): print(1) elif(len(m['m'])== 2 and (abs(m['m'][0]-m['m'][1]) == 1 or abs(m['m'][0]-m['m'][1])==2)): print(1) else: print(2) if __name__ =="__main__": if(file): if path.exists('input.txt'): sys.stdin=open('input.txt', 'r') sys.stdout=open('output.txt','w') else: input=sys.stdin.readline solve() ```
instruction
0
5,114
19
10,228
Yes
output
1
5,114
19
10,229
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Tokitsukaze is playing a game derivated from Japanese mahjong. In this game, she has three tiles in her hand. Each tile she owns is a suited tile, which means it has a suit (manzu, pinzu or souzu) and a number (a digit ranged from 1 to 9). In this problem, we use one digit and one lowercase letter, which is the first character of the suit, to represent a suited tile. All possible suited tiles are represented as 1m, 2m, …, 9m, 1p, 2p, …, 9p, 1s, 2s, …, 9s. In order to win the game, she must have at least one mentsu (described below) in her hand, so sometimes she should draw extra suited tiles. After drawing a tile, the number of her tiles increases by one. She can draw any tiles she wants, including those already in her hand. Do you know the minimum number of extra suited tiles she needs to draw so that she can win? Here are some useful definitions in this game: * A mentsu, also known as meld, is formed by a koutsu or a shuntsu; * A koutsu, also known as triplet, is made of three identical tiles, such as [1m, 1m, 1m], however, [1m, 1p, 1s] or [1m, 4m, 7m] is NOT a koutsu; * A shuntsu, also known as sequence, is made of three sequential numbered tiles in the same suit, such as [1m, 2m, 3m] and [5s, 7s, 6s], however, [9m, 1m, 2m] or [1m, 2p, 3s] is NOT a shuntsu. Some examples: * [2m, 3p, 2s, 4m, 1s, 2s, 4s] β€” it contains no koutsu or shuntsu, so it includes no mentsu; * [4s, 3m, 3p, 4s, 5p, 4s, 5p] β€” it contains a koutsu, [4s, 4s, 4s], but no shuntsu, so it includes a mentsu; * [5p, 5s, 9m, 4p, 1s, 7p, 7m, 6p] β€” it contains no koutsu but a shuntsu, [5p, 4p, 6p] or [5p, 7p, 6p], so it includes a mentsu. Note that the order of tiles is unnecessary and you can assume the number of each type of suited tiles she can draw is infinite. Input The only line contains three strings β€” the tiles in Tokitsukaze's hand. For each string, the first character is a digit ranged from 1 to 9 and the second character is m, p or s. Output Print a single integer β€” the minimum number of extra suited tiles she needs to draw. Examples Input 1s 2s 3s Output 0 Input 9m 9m 9m Output 0 Input 3p 9m 2p Output 1 Note In the first example, Tokitsukaze already has a shuntsu. In the second example, Tokitsukaze already has a koutsu. In the third example, Tokitsukaze can get a shuntsu by drawing one suited tile β€” 1p or 4p. The resulting tiles will be [3p, 9m, 2p, 1p] or [3p, 9m, 2p, 4p]. Submitted Solution: ``` s=input().split() l=[] s.sort() for i in s: l.append(int(i[0])) l.append(i[1]) if l[0]==l[2] and l[1]==l[3]: if l[0]==l[4] and l[1]==l[5]: print(0) else: print(1) elif l[4]==l[2] and l[5]==l[3]: print(1) elif l[0]+1==l[2] and l[1]==l[3]: if l[2]+1==l[4] and l[3]==l[5]: print(0) else: print(1) elif l[2]+1==l[4] and l[5]==l[3]: print(1) elif l[0]+2==l[4] and l[1]==l[5]: print(1) else: print(2) ```
instruction
0
5,115
19
10,230
No
output
1
5,115
19
10,231
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Tokitsukaze is playing a game derivated from Japanese mahjong. In this game, she has three tiles in her hand. Each tile she owns is a suited tile, which means it has a suit (manzu, pinzu or souzu) and a number (a digit ranged from 1 to 9). In this problem, we use one digit and one lowercase letter, which is the first character of the suit, to represent a suited tile. All possible suited tiles are represented as 1m, 2m, …, 9m, 1p, 2p, …, 9p, 1s, 2s, …, 9s. In order to win the game, she must have at least one mentsu (described below) in her hand, so sometimes she should draw extra suited tiles. After drawing a tile, the number of her tiles increases by one. She can draw any tiles she wants, including those already in her hand. Do you know the minimum number of extra suited tiles she needs to draw so that she can win? Here are some useful definitions in this game: * A mentsu, also known as meld, is formed by a koutsu or a shuntsu; * A koutsu, also known as triplet, is made of three identical tiles, such as [1m, 1m, 1m], however, [1m, 1p, 1s] or [1m, 4m, 7m] is NOT a koutsu; * A shuntsu, also known as sequence, is made of three sequential numbered tiles in the same suit, such as [1m, 2m, 3m] and [5s, 7s, 6s], however, [9m, 1m, 2m] or [1m, 2p, 3s] is NOT a shuntsu. Some examples: * [2m, 3p, 2s, 4m, 1s, 2s, 4s] β€” it contains no koutsu or shuntsu, so it includes no mentsu; * [4s, 3m, 3p, 4s, 5p, 4s, 5p] β€” it contains a koutsu, [4s, 4s, 4s], but no shuntsu, so it includes a mentsu; * [5p, 5s, 9m, 4p, 1s, 7p, 7m, 6p] β€” it contains no koutsu but a shuntsu, [5p, 4p, 6p] or [5p, 7p, 6p], so it includes a mentsu. Note that the order of tiles is unnecessary and you can assume the number of each type of suited tiles she can draw is infinite. Input The only line contains three strings β€” the tiles in Tokitsukaze's hand. For each string, the first character is a digit ranged from 1 to 9 and the second character is m, p or s. Output Print a single integer β€” the minimum number of extra suited tiles she needs to draw. Examples Input 1s 2s 3s Output 0 Input 9m 9m 9m Output 0 Input 3p 9m 2p Output 1 Note In the first example, Tokitsukaze already has a shuntsu. In the second example, Tokitsukaze already has a koutsu. In the third example, Tokitsukaze can get a shuntsu by drawing one suited tile β€” 1p or 4p. The resulting tiles will be [3p, 9m, 2p, 1p] or [3p, 9m, 2p, 4p]. Submitted Solution: ``` def process1(n): Dicts = {1:'0 A', 2:'1 B', 3:'2 A', 0:'1 A'} mod = n%4 return Dicts[mod] def tile3(n): list = n.split(" ") list = sorted(list) print(list) list2 = [] for i in range(len(list)): count = 0 for j in range(len(list)): if list[i]==list[j]: count+=1 list2.append(count) if max(list2)>=3: # Trường hợp cΓ³ 3 tile giα»‘ng nhau return 0 elif max(list2)==2 and min(list2)>=1: # Trường hợp cΓ³ 2 tile giα»‘ng nhau return 1 elif list2.count(1)>=3: # Trường hợp cαΊ£ 3 tile khΓ‘c nhau for i in list: if int(list[0][0])+2==int(list[1][0])+1==int(list[2][0]) and list[0][1]==list[1][1]==list[2][1]: return 0 elif (list[0][1]!=list[1][1] and list[1][1]!=list[2][1] and list[2][1] != list[0][1]) or \ (int(list[1][0])-int(list[0][0])>2) and (int(list[2][0])-int(list[1][0])>2): return 2 elif (list[0][1]==list[1][1] and list[1][1]!=list[2][1] and int(list[0][0])+1==int(list[1][0])) or \ (list[1][1]==list[2][1] and list[1][1]!=list[0][1] and int(list[1][0])+1==int(list[2][0])): return 1 else: return 1 n = input() s = tile3(n) print(s) ```
instruction
0
5,116
19
10,232
No
output
1
5,116
19
10,233
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Tokitsukaze is playing a game derivated from Japanese mahjong. In this game, she has three tiles in her hand. Each tile she owns is a suited tile, which means it has a suit (manzu, pinzu or souzu) and a number (a digit ranged from 1 to 9). In this problem, we use one digit and one lowercase letter, which is the first character of the suit, to represent a suited tile. All possible suited tiles are represented as 1m, 2m, …, 9m, 1p, 2p, …, 9p, 1s, 2s, …, 9s. In order to win the game, she must have at least one mentsu (described below) in her hand, so sometimes she should draw extra suited tiles. After drawing a tile, the number of her tiles increases by one. She can draw any tiles she wants, including those already in her hand. Do you know the minimum number of extra suited tiles she needs to draw so that she can win? Here are some useful definitions in this game: * A mentsu, also known as meld, is formed by a koutsu or a shuntsu; * A koutsu, also known as triplet, is made of three identical tiles, such as [1m, 1m, 1m], however, [1m, 1p, 1s] or [1m, 4m, 7m] is NOT a koutsu; * A shuntsu, also known as sequence, is made of three sequential numbered tiles in the same suit, such as [1m, 2m, 3m] and [5s, 7s, 6s], however, [9m, 1m, 2m] or [1m, 2p, 3s] is NOT a shuntsu. Some examples: * [2m, 3p, 2s, 4m, 1s, 2s, 4s] β€” it contains no koutsu or shuntsu, so it includes no mentsu; * [4s, 3m, 3p, 4s, 5p, 4s, 5p] β€” it contains a koutsu, [4s, 4s, 4s], but no shuntsu, so it includes a mentsu; * [5p, 5s, 9m, 4p, 1s, 7p, 7m, 6p] β€” it contains no koutsu but a shuntsu, [5p, 4p, 6p] or [5p, 7p, 6p], so it includes a mentsu. Note that the order of tiles is unnecessary and you can assume the number of each type of suited tiles she can draw is infinite. Input The only line contains three strings β€” the tiles in Tokitsukaze's hand. For each string, the first character is a digit ranged from 1 to 9 and the second character is m, p or s. Output Print a single integer β€” the minimum number of extra suited tiles she needs to draw. Examples Input 1s 2s 3s Output 0 Input 9m 9m 9m Output 0 Input 3p 9m 2p Output 1 Note In the first example, Tokitsukaze already has a shuntsu. In the second example, Tokitsukaze already has a koutsu. In the third example, Tokitsukaze can get a shuntsu by drawing one suited tile β€” 1p or 4p. The resulting tiles will be [3p, 9m, 2p, 1p] or [3p, 9m, 2p, 4p]. Submitted Solution: ``` arr = list(input().split()) x = arr[0] y = arr[1] z = arr[2] if x==y==z: print(0) elif x[1] == y[1] == z[1]: arr = [] arr.append(int(x[0])) arr.append(int(y[0])) arr.append(int(z[0])) arr.sort() #print(arr) if arr[1]-arr[0]==arr[2]-arr[1]==1: print(0) elif arr[1]-arr[0]==1: print(1) elif arr[2]-arr[1]==1: print(1) elif arr[2]-arr[0]==1: print(1) else: print(2) elif x==y or y==z or z==x: print(1) elif x[1]==y[1] and abs(int(x[0])-int(y[0]))==1: print(1) elif y[1]==z[1] and abs(int(y[0])-int(z[0]))==1: print(1) elif z[1]==x[1] and abs(int(z[0])-int(x[0]))==1: print(1) else: print(2) ```
instruction
0
5,117
19
10,234
No
output
1
5,117
19
10,235
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Tokitsukaze is playing a game derivated from Japanese mahjong. In this game, she has three tiles in her hand. Each tile she owns is a suited tile, which means it has a suit (manzu, pinzu or souzu) and a number (a digit ranged from 1 to 9). In this problem, we use one digit and one lowercase letter, which is the first character of the suit, to represent a suited tile. All possible suited tiles are represented as 1m, 2m, …, 9m, 1p, 2p, …, 9p, 1s, 2s, …, 9s. In order to win the game, she must have at least one mentsu (described below) in her hand, so sometimes she should draw extra suited tiles. After drawing a tile, the number of her tiles increases by one. She can draw any tiles she wants, including those already in her hand. Do you know the minimum number of extra suited tiles she needs to draw so that she can win? Here are some useful definitions in this game: * A mentsu, also known as meld, is formed by a koutsu or a shuntsu; * A koutsu, also known as triplet, is made of three identical tiles, such as [1m, 1m, 1m], however, [1m, 1p, 1s] or [1m, 4m, 7m] is NOT a koutsu; * A shuntsu, also known as sequence, is made of three sequential numbered tiles in the same suit, such as [1m, 2m, 3m] and [5s, 7s, 6s], however, [9m, 1m, 2m] or [1m, 2p, 3s] is NOT a shuntsu. Some examples: * [2m, 3p, 2s, 4m, 1s, 2s, 4s] β€” it contains no koutsu or shuntsu, so it includes no mentsu; * [4s, 3m, 3p, 4s, 5p, 4s, 5p] β€” it contains a koutsu, [4s, 4s, 4s], but no shuntsu, so it includes a mentsu; * [5p, 5s, 9m, 4p, 1s, 7p, 7m, 6p] β€” it contains no koutsu but a shuntsu, [5p, 4p, 6p] or [5p, 7p, 6p], so it includes a mentsu. Note that the order of tiles is unnecessary and you can assume the number of each type of suited tiles she can draw is infinite. Input The only line contains three strings β€” the tiles in Tokitsukaze's hand. For each string, the first character is a digit ranged from 1 to 9 and the second character is m, p or s. Output Print a single integer β€” the minimum number of extra suited tiles she needs to draw. Examples Input 1s 2s 3s Output 0 Input 9m 9m 9m Output 0 Input 3p 9m 2p Output 1 Note In the first example, Tokitsukaze already has a shuntsu. In the second example, Tokitsukaze already has a koutsu. In the third example, Tokitsukaze can get a shuntsu by drawing one suited tile β€” 1p or 4p. The resulting tiles will be [3p, 9m, 2p, 1p] or [3p, 9m, 2p, 4p]. Submitted Solution: ``` a,b,c = map(str, input().split()) a = a[1]+a[0] b = b[1]+b[0] c = c[1]+c[0] n = [a,b,c] m = [0]*9 p = [0]*9 s = [0]*9 for i in range(3): if(n[i][0]=="m"): m[int(n[i][1])-1]+=1 if(n[i][0]=="p"): p[int(n[i][1])-1]+=1 if(n[i][0]=="s"): s[int(n[i][1])-1]+=1 ans = 2 #print(m,p,s) for i in range(9): if(3-m[i]<ans): ans = 3-m[i] if(0<i and i<8): if(3-m[i-1]-m[i]-m[i+1]<ans): ans = 3-m[i-1]-m[i]-m[i+1] if(3-s[i]<ans): ans = 3-s[i] if(0<i and i<8): if(3-s[i-1]-s[i]-s[i+1]<ans): #print(23) ans = 3-s[i-1]-s[i]-s[i+1] #print(ans, 9) if(3-p[i]<ans): ans = 3-p[i] #print(3-s[i-1]-s[i]-s[i+1]) if(0<i and i<8): if(3-p[i-1]-p[i]-p[i+1]<ans): ans = 3-p[i-1]-p[i]-p[i+1] print(ans) ```
instruction
0
5,118
19
10,236
No
output
1
5,118
19
10,237
Provide tags and a correct Python 3 solution for this coding contest problem. Welcome to Rockport City! It is time for your first ever race in the game against Ronnie. To make the race interesting, you have bet a dollars and Ronnie has bet b dollars. But the fans seem to be disappointed. The excitement of the fans is given by gcd(a,b), where gcd(x, y) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers x and y. To make the race more exciting, you can perform two types of operations: 1. Increase both a and b by 1. 2. Decrease both a and b by 1. This operation can only be performed if both a and b are greater than 0. In one move, you can perform any one of these operations. You can perform arbitrary (possibly zero) number of moves. Determine the maximum excitement the fans can get and the minimum number of moves required to achieve it. Note that gcd(x,0)=x for any x β‰₯ 0. Input The first line of input contains a single integer t (1≀ t≀ 5β‹… 10^3) β€” the number of test cases. The first and the only line of each test case contains two integers a and b (0≀ a, b≀ 10^{18}). Output For each test case, print a single line containing two integers. If the fans can get infinite excitement, print 0 0. Otherwise, the first integer must be the maximum excitement the fans can get, and the second integer must be the minimum number of moves required to achieve that excitement. Example Input 4 8 5 1 2 4 4 3 9 Output 3 1 1 0 0 0 6 3 Note For the first test case, you can apply the first operation 1 time to get a=9 and b=6. It can be shown that 3 is the maximum excitement possible. For the second test case, no matter how many operations you apply, the fans will always have an excitement equal to 1. Since the initial excitement is also 1, you don't need to apply any operation. For the third case, the fans can get infinite excitement by applying the first operation an infinite amount of times. For the fourth test case, you can apply the second operation 3 times to get a=0 and b=6. Since, gcd(0,6)=6, the fans will get an excitement of 6.
instruction
0
5,269
19
10,538
Tags: greedy, math, number theory Correct Solution: ``` import sys input=sys.stdin.readline for _ in range(int(input())): a,b=map(int,input().split()) if(a==b): print(0,0) else: dif=abs(a-b) mi=min(a,b) rem=mi%dif print(dif,min(rem,abs(dif-rem))) ```
output
1
5,269
19
10,539
Provide tags and a correct Python 3 solution for this coding contest problem. Welcome to Rockport City! It is time for your first ever race in the game against Ronnie. To make the race interesting, you have bet a dollars and Ronnie has bet b dollars. But the fans seem to be disappointed. The excitement of the fans is given by gcd(a,b), where gcd(x, y) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers x and y. To make the race more exciting, you can perform two types of operations: 1. Increase both a and b by 1. 2. Decrease both a and b by 1. This operation can only be performed if both a and b are greater than 0. In one move, you can perform any one of these operations. You can perform arbitrary (possibly zero) number of moves. Determine the maximum excitement the fans can get and the minimum number of moves required to achieve it. Note that gcd(x,0)=x for any x β‰₯ 0. Input The first line of input contains a single integer t (1≀ t≀ 5β‹… 10^3) β€” the number of test cases. The first and the only line of each test case contains two integers a and b (0≀ a, b≀ 10^{18}). Output For each test case, print a single line containing two integers. If the fans can get infinite excitement, print 0 0. Otherwise, the first integer must be the maximum excitement the fans can get, and the second integer must be the minimum number of moves required to achieve that excitement. Example Input 4 8 5 1 2 4 4 3 9 Output 3 1 1 0 0 0 6 3 Note For the first test case, you can apply the first operation 1 time to get a=9 and b=6. It can be shown that 3 is the maximum excitement possible. For the second test case, no matter how many operations you apply, the fans will always have an excitement equal to 1. Since the initial excitement is also 1, you don't need to apply any operation. For the third case, the fans can get infinite excitement by applying the first operation an infinite amount of times. For the fourth test case, you can apply the second operation 3 times to get a=0 and b=6. Since, gcd(0,6)=6, the fans will get an excitement of 6.
instruction
0
5,270
19
10,540
Tags: greedy, math, number theory Correct Solution: ``` import sys import math #sys.stdin=open('input.txt','r') #sys.stdout=open('output.txt','w') def solve(): #n=int(input()) a,b=map(int,input().split()) if(a==b): print("0 0") else: r=abs(a-b) if(math.gcd(a,b)==r): print(r,0) else: e=a//r y1=((e+1)*r)-a y2=abs(((e)*r)-a) print(abs(a-b),min(y1,y2)) t=int(input()) while(t!=0): solve() t-=1 ```
output
1
5,270
19
10,541
Provide tags and a correct Python 3 solution for this coding contest problem. Welcome to Rockport City! It is time for your first ever race in the game against Ronnie. To make the race interesting, you have bet a dollars and Ronnie has bet b dollars. But the fans seem to be disappointed. The excitement of the fans is given by gcd(a,b), where gcd(x, y) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers x and y. To make the race more exciting, you can perform two types of operations: 1. Increase both a and b by 1. 2. Decrease both a and b by 1. This operation can only be performed if both a and b are greater than 0. In one move, you can perform any one of these operations. You can perform arbitrary (possibly zero) number of moves. Determine the maximum excitement the fans can get and the minimum number of moves required to achieve it. Note that gcd(x,0)=x for any x β‰₯ 0. Input The first line of input contains a single integer t (1≀ t≀ 5β‹… 10^3) β€” the number of test cases. The first and the only line of each test case contains two integers a and b (0≀ a, b≀ 10^{18}). Output For each test case, print a single line containing two integers. If the fans can get infinite excitement, print 0 0. Otherwise, the first integer must be the maximum excitement the fans can get, and the second integer must be the minimum number of moves required to achieve that excitement. Example Input 4 8 5 1 2 4 4 3 9 Output 3 1 1 0 0 0 6 3 Note For the first test case, you can apply the first operation 1 time to get a=9 and b=6. It can be shown that 3 is the maximum excitement possible. For the second test case, no matter how many operations you apply, the fans will always have an excitement equal to 1. Since the initial excitement is also 1, you don't need to apply any operation. For the third case, the fans can get infinite excitement by applying the first operation an infinite amount of times. For the fourth test case, you can apply the second operation 3 times to get a=0 and b=6. Since, gcd(0,6)=6, the fans will get an excitement of 6.
instruction
0
5,271
19
10,542
Tags: greedy, math, number theory Correct Solution: ``` from math import * t=int(input()) for i in range(t): temp=input().split() x=int(temp[0]);y=int(temp[1]) gre=abs(x-y) if(gre!=0): num = min(x%gre,y%gre,gre-x%gre,gre-x%gre) else: num = 0 print(str(gre)+" "+str(num)) ```
output
1
5,271
19
10,543
Provide tags and a correct Python 3 solution for this coding contest problem. Welcome to Rockport City! It is time for your first ever race in the game against Ronnie. To make the race interesting, you have bet a dollars and Ronnie has bet b dollars. But the fans seem to be disappointed. The excitement of the fans is given by gcd(a,b), where gcd(x, y) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers x and y. To make the race more exciting, you can perform two types of operations: 1. Increase both a and b by 1. 2. Decrease both a and b by 1. This operation can only be performed if both a and b are greater than 0. In one move, you can perform any one of these operations. You can perform arbitrary (possibly zero) number of moves. Determine the maximum excitement the fans can get and the minimum number of moves required to achieve it. Note that gcd(x,0)=x for any x β‰₯ 0. Input The first line of input contains a single integer t (1≀ t≀ 5β‹… 10^3) β€” the number of test cases. The first and the only line of each test case contains two integers a and b (0≀ a, b≀ 10^{18}). Output For each test case, print a single line containing two integers. If the fans can get infinite excitement, print 0 0. Otherwise, the first integer must be the maximum excitement the fans can get, and the second integer must be the minimum number of moves required to achieve that excitement. Example Input 4 8 5 1 2 4 4 3 9 Output 3 1 1 0 0 0 6 3 Note For the first test case, you can apply the first operation 1 time to get a=9 and b=6. It can be shown that 3 is the maximum excitement possible. For the second test case, no matter how many operations you apply, the fans will always have an excitement equal to 1. Since the initial excitement is also 1, you don't need to apply any operation. For the third case, the fans can get infinite excitement by applying the first operation an infinite amount of times. For the fourth test case, you can apply the second operation 3 times to get a=0 and b=6. Since, gcd(0,6)=6, the fans will get an excitement of 6.
instruction
0
5,272
19
10,544
Tags: greedy, math, number theory Correct Solution: ``` for _ in range(int(input())): a,b=map(int,input().split()) x=min(a,b) y=max(a,b) if x==y: print("0 0") else: print(y-x, end=" ") b=y-x a=x%b print(min(a, b-a)) ```
output
1
5,272
19
10,545
Provide tags and a correct Python 3 solution for this coding contest problem. Welcome to Rockport City! It is time for your first ever race in the game against Ronnie. To make the race interesting, you have bet a dollars and Ronnie has bet b dollars. But the fans seem to be disappointed. The excitement of the fans is given by gcd(a,b), where gcd(x, y) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers x and y. To make the race more exciting, you can perform two types of operations: 1. Increase both a and b by 1. 2. Decrease both a and b by 1. This operation can only be performed if both a and b are greater than 0. In one move, you can perform any one of these operations. You can perform arbitrary (possibly zero) number of moves. Determine the maximum excitement the fans can get and the minimum number of moves required to achieve it. Note that gcd(x,0)=x for any x β‰₯ 0. Input The first line of input contains a single integer t (1≀ t≀ 5β‹… 10^3) β€” the number of test cases. The first and the only line of each test case contains two integers a and b (0≀ a, b≀ 10^{18}). Output For each test case, print a single line containing two integers. If the fans can get infinite excitement, print 0 0. Otherwise, the first integer must be the maximum excitement the fans can get, and the second integer must be the minimum number of moves required to achieve that excitement. Example Input 4 8 5 1 2 4 4 3 9 Output 3 1 1 0 0 0 6 3 Note For the first test case, you can apply the first operation 1 time to get a=9 and b=6. It can be shown that 3 is the maximum excitement possible. For the second test case, no matter how many operations you apply, the fans will always have an excitement equal to 1. Since the initial excitement is also 1, you don't need to apply any operation. For the third case, the fans can get infinite excitement by applying the first operation an infinite amount of times. For the fourth test case, you can apply the second operation 3 times to get a=0 and b=6. Since, gcd(0,6)=6, the fans will get an excitement of 6.
instruction
0
5,273
19
10,546
Tags: greedy, math, number theory Correct Solution: ``` t = int(input()) for _ in range(t): a, b = map(int,input().split()) if(a == b): print('0 0') elif(abs(a-b) == 1): print('1 0') elif(a==0 and b!=0): print(str(b)+" 0") elif(b==0 and a!=0): print(str(a)+" 0") else: exc = abs(a-b) if(a < b): k = 0 while(k < a): pre = k k = k + exc if((a-pre) > (k-a)): print(str(exc)+" "+str(k-a)) else: print(str(exc)+" "+str(a-pre)) else: k = 0 while(k < b): pre = k k = k + exc if((b-pre) > (k-b)): print(str(exc)+" "+str(k-b)) else: print(str(exc)+" "+str(b-pre)) ```
output
1
5,273
19
10,547
Provide tags and a correct Python 3 solution for this coding contest problem. Welcome to Rockport City! It is time for your first ever race in the game against Ronnie. To make the race interesting, you have bet a dollars and Ronnie has bet b dollars. But the fans seem to be disappointed. The excitement of the fans is given by gcd(a,b), where gcd(x, y) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers x and y. To make the race more exciting, you can perform two types of operations: 1. Increase both a and b by 1. 2. Decrease both a and b by 1. This operation can only be performed if both a and b are greater than 0. In one move, you can perform any one of these operations. You can perform arbitrary (possibly zero) number of moves. Determine the maximum excitement the fans can get and the minimum number of moves required to achieve it. Note that gcd(x,0)=x for any x β‰₯ 0. Input The first line of input contains a single integer t (1≀ t≀ 5β‹… 10^3) β€” the number of test cases. The first and the only line of each test case contains two integers a and b (0≀ a, b≀ 10^{18}). Output For each test case, print a single line containing two integers. If the fans can get infinite excitement, print 0 0. Otherwise, the first integer must be the maximum excitement the fans can get, and the second integer must be the minimum number of moves required to achieve that excitement. Example Input 4 8 5 1 2 4 4 3 9 Output 3 1 1 0 0 0 6 3 Note For the first test case, you can apply the first operation 1 time to get a=9 and b=6. It can be shown that 3 is the maximum excitement possible. For the second test case, no matter how many operations you apply, the fans will always have an excitement equal to 1. Since the initial excitement is also 1, you don't need to apply any operation. For the third case, the fans can get infinite excitement by applying the first operation an infinite amount of times. For the fourth test case, you can apply the second operation 3 times to get a=0 and b=6. Since, gcd(0,6)=6, the fans will get an excitement of 6.
instruction
0
5,274
19
10,548
Tags: greedy, math, number theory Correct Solution: ``` t=int(input()) for i in range (t): a,b=map(int,input().strip().split()) if a==b: print(0,0) elif abs(a-b)==1: print(1,0) else: gcd=abs(a-b) minimum=min(a,b) temp=minimum//gcd prv=(minimum-(temp*gcd)) nxt=((temp+1)*gcd)-minimum print(gcd,min(nxt,prv)) ```
output
1
5,274
19
10,549
Provide tags and a correct Python 3 solution for this coding contest problem. Welcome to Rockport City! It is time for your first ever race in the game against Ronnie. To make the race interesting, you have bet a dollars and Ronnie has bet b dollars. But the fans seem to be disappointed. The excitement of the fans is given by gcd(a,b), where gcd(x, y) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers x and y. To make the race more exciting, you can perform two types of operations: 1. Increase both a and b by 1. 2. Decrease both a and b by 1. This operation can only be performed if both a and b are greater than 0. In one move, you can perform any one of these operations. You can perform arbitrary (possibly zero) number of moves. Determine the maximum excitement the fans can get and the minimum number of moves required to achieve it. Note that gcd(x,0)=x for any x β‰₯ 0. Input The first line of input contains a single integer t (1≀ t≀ 5β‹… 10^3) β€” the number of test cases. The first and the only line of each test case contains two integers a and b (0≀ a, b≀ 10^{18}). Output For each test case, print a single line containing two integers. If the fans can get infinite excitement, print 0 0. Otherwise, the first integer must be the maximum excitement the fans can get, and the second integer must be the minimum number of moves required to achieve that excitement. Example Input 4 8 5 1 2 4 4 3 9 Output 3 1 1 0 0 0 6 3 Note For the first test case, you can apply the first operation 1 time to get a=9 and b=6. It can be shown that 3 is the maximum excitement possible. For the second test case, no matter how many operations you apply, the fans will always have an excitement equal to 1. Since the initial excitement is also 1, you don't need to apply any operation. For the third case, the fans can get infinite excitement by applying the first operation an infinite amount of times. For the fourth test case, you can apply the second operation 3 times to get a=0 and b=6. Since, gcd(0,6)=6, the fans will get an excitement of 6.
instruction
0
5,275
19
10,550
Tags: greedy, math, number theory Correct Solution: ``` from sys import stdin input = stdin.readline t = int(input()) for _ in range(t): a, b = [int(x) for x in input().split()] d = abs(a - b) if d == 0: print(0, 0) continue minus_count = a % d plus_count = d - minus_count if min(a, b) - minus_count < 0: print(d, plus_count) continue if minus_count < plus_count: print(d, minus_count) else: print(d, plus_count) ```
output
1
5,275
19
10,551
Provide tags and a correct Python 3 solution for this coding contest problem. Welcome to Rockport City! It is time for your first ever race in the game against Ronnie. To make the race interesting, you have bet a dollars and Ronnie has bet b dollars. But the fans seem to be disappointed. The excitement of the fans is given by gcd(a,b), where gcd(x, y) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers x and y. To make the race more exciting, you can perform two types of operations: 1. Increase both a and b by 1. 2. Decrease both a and b by 1. This operation can only be performed if both a and b are greater than 0. In one move, you can perform any one of these operations. You can perform arbitrary (possibly zero) number of moves. Determine the maximum excitement the fans can get and the minimum number of moves required to achieve it. Note that gcd(x,0)=x for any x β‰₯ 0. Input The first line of input contains a single integer t (1≀ t≀ 5β‹… 10^3) β€” the number of test cases. The first and the only line of each test case contains two integers a and b (0≀ a, b≀ 10^{18}). Output For each test case, print a single line containing two integers. If the fans can get infinite excitement, print 0 0. Otherwise, the first integer must be the maximum excitement the fans can get, and the second integer must be the minimum number of moves required to achieve that excitement. Example Input 4 8 5 1 2 4 4 3 9 Output 3 1 1 0 0 0 6 3 Note For the first test case, you can apply the first operation 1 time to get a=9 and b=6. It can be shown that 3 is the maximum excitement possible. For the second test case, no matter how many operations you apply, the fans will always have an excitement equal to 1. Since the initial excitement is also 1, you don't need to apply any operation. For the third case, the fans can get infinite excitement by applying the first operation an infinite amount of times. For the fourth test case, you can apply the second operation 3 times to get a=0 and b=6. Since, gcd(0,6)=6, the fans will get an excitement of 6.
instruction
0
5,276
19
10,552
Tags: greedy, math, number theory Correct Solution: ``` from math import ceil for i in range(int(input())): a,b=map(int,input().split()) if a==b: print(0,0) else: mgcd=max(a,b)-min(a,b) temp=min(a,b) x=ceil(temp/mgcd) z=temp//mgcd y=(x*mgcd)-temp y2=temp-(z*mgcd) ans=min(temp,y,y2) #print(temp,mgcd,x,y) print(mgcd,ans) ```
output
1
5,276
19
10,553
Provide tags and a correct Python 3 solution for this coding contest problem. Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1. You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all already placed slimes. Then, while the last two slimes in the row have the same value v, you combine them together to create a slime with value v + 1. You would like to see what the final state of the row is after you've added all n slimes. Please print the values of the slimes in the row from left to right. Input The first line of the input will contain a single integer, n (1 ≀ n ≀ 100 000). Output Output a single line with k integers, where k is the number of slimes in the row after you've finished the procedure described in the problem statement. The i-th of these numbers should be the value of the i-th slime from the left. Examples Input 1 Output 1 Input 2 Output 2 Input 3 Output 2 1 Input 8 Output 4 Note In the first sample, we only have a single slime with value 1. The final state of the board is just a single slime with value 1. In the second sample, we perform the following steps: Initially we place a single slime in a row by itself. Thus, row is initially 1. Then, we will add another slime. The row is now 1 1. Since two rightmost slimes have the same values, we should replace these slimes with one with value 2. Thus, the final state of the board is 2. In the third sample, after adding the first two slimes, our row is 2. After adding one more slime, the row becomes 2 1. In the last sample, the steps look as follows: 1. 1 2. 2 3. 2 1 4. 3 5. 3 1 6. 3 2 7. 3 2 1 8. 4
instruction
0
5,441
19
10,882
Tags: implementation Correct Solution: ``` n=int(input()) for i in range(20,-1,-1): if(2**i<=n): print(i+1,end=' ') n-=2**i ```
output
1
5,441
19
10,883
Provide tags and a correct Python 3 solution for this coding contest problem. Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1. You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all already placed slimes. Then, while the last two slimes in the row have the same value v, you combine them together to create a slime with value v + 1. You would like to see what the final state of the row is after you've added all n slimes. Please print the values of the slimes in the row from left to right. Input The first line of the input will contain a single integer, n (1 ≀ n ≀ 100 000). Output Output a single line with k integers, where k is the number of slimes in the row after you've finished the procedure described in the problem statement. The i-th of these numbers should be the value of the i-th slime from the left. Examples Input 1 Output 1 Input 2 Output 2 Input 3 Output 2 1 Input 8 Output 4 Note In the first sample, we only have a single slime with value 1. The final state of the board is just a single slime with value 1. In the second sample, we perform the following steps: Initially we place a single slime in a row by itself. Thus, row is initially 1. Then, we will add another slime. The row is now 1 1. Since two rightmost slimes have the same values, we should replace these slimes with one with value 2. Thus, the final state of the board is 2. In the third sample, after adding the first two slimes, our row is 2. After adding one more slime, the row becomes 2 1. In the last sample, the steps look as follows: 1. 1 2. 2 3. 2 1 4. 3 5. 3 1 6. 3 2 7. 3 2 1 8. 4
instruction
0
5,442
19
10,884
Tags: implementation Correct Solution: ``` n = bin(int(input()))[2:] l = len(n) for i in n: if i == "1": print(l, end = " ") l -= 1 ```
output
1
5,442
19
10,885
Provide tags and a correct Python 3 solution for this coding contest problem. Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1. You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all already placed slimes. Then, while the last two slimes in the row have the same value v, you combine them together to create a slime with value v + 1. You would like to see what the final state of the row is after you've added all n slimes. Please print the values of the slimes in the row from left to right. Input The first line of the input will contain a single integer, n (1 ≀ n ≀ 100 000). Output Output a single line with k integers, where k is the number of slimes in the row after you've finished the procedure described in the problem statement. The i-th of these numbers should be the value of the i-th slime from the left. Examples Input 1 Output 1 Input 2 Output 2 Input 3 Output 2 1 Input 8 Output 4 Note In the first sample, we only have a single slime with value 1. The final state of the board is just a single slime with value 1. In the second sample, we perform the following steps: Initially we place a single slime in a row by itself. Thus, row is initially 1. Then, we will add another slime. The row is now 1 1. Since two rightmost slimes have the same values, we should replace these slimes with one with value 2. Thus, the final state of the board is 2. In the third sample, after adding the first two slimes, our row is 2. After adding one more slime, the row becomes 2 1. In the last sample, the steps look as follows: 1. 1 2. 2 3. 2 1 4. 3 5. 3 1 6. 3 2 7. 3 2 1 8. 4
instruction
0
5,443
19
10,886
Tags: implementation Correct Solution: ``` L=[1]; for i in range(2,int(input())+1): L.append(1); a=len(L)-1; while(a>0): if L[a]==L[a-1]: L[a-1]=L[a]+1; L[a:a+1]=[]; else: break a=len(L)-1; if a==0: break for i in L: print(i,end=' ') ```
output
1
5,443
19
10,887
Provide tags and a correct Python 3 solution for this coding contest problem. Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1. You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all already placed slimes. Then, while the last two slimes in the row have the same value v, you combine them together to create a slime with value v + 1. You would like to see what the final state of the row is after you've added all n slimes. Please print the values of the slimes in the row from left to right. Input The first line of the input will contain a single integer, n (1 ≀ n ≀ 100 000). Output Output a single line with k integers, where k is the number of slimes in the row after you've finished the procedure described in the problem statement. The i-th of these numbers should be the value of the i-th slime from the left. Examples Input 1 Output 1 Input 2 Output 2 Input 3 Output 2 1 Input 8 Output 4 Note In the first sample, we only have a single slime with value 1. The final state of the board is just a single slime with value 1. In the second sample, we perform the following steps: Initially we place a single slime in a row by itself. Thus, row is initially 1. Then, we will add another slime. The row is now 1 1. Since two rightmost slimes have the same values, we should replace these slimes with one with value 2. Thus, the final state of the board is 2. In the third sample, after adding the first two slimes, our row is 2. After adding one more slime, the row becomes 2 1. In the last sample, the steps look as follows: 1. 1 2. 2 3. 2 1 4. 3 5. 3 1 6. 3 2 7. 3 2 1 8. 4
instruction
0
5,444
19
10,888
Tags: implementation Correct Solution: ``` n = int(input()) l = "{:b}".format(n) r = [] for i, c in enumerate(l): if c=="1": r += [str(len(l)-i)] print(" ".join(r)) ```
output
1
5,444
19
10,889
Provide tags and a correct Python 3 solution for this coding contest problem. Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1. You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all already placed slimes. Then, while the last two slimes in the row have the same value v, you combine them together to create a slime with value v + 1. You would like to see what the final state of the row is after you've added all n slimes. Please print the values of the slimes in the row from left to right. Input The first line of the input will contain a single integer, n (1 ≀ n ≀ 100 000). Output Output a single line with k integers, where k is the number of slimes in the row after you've finished the procedure described in the problem statement. The i-th of these numbers should be the value of the i-th slime from the left. Examples Input 1 Output 1 Input 2 Output 2 Input 3 Output 2 1 Input 8 Output 4 Note In the first sample, we only have a single slime with value 1. The final state of the board is just a single slime with value 1. In the second sample, we perform the following steps: Initially we place a single slime in a row by itself. Thus, row is initially 1. Then, we will add another slime. The row is now 1 1. Since two rightmost slimes have the same values, we should replace these slimes with one with value 2. Thus, the final state of the board is 2. In the third sample, after adding the first two slimes, our row is 2. After adding one more slime, the row becomes 2 1. In the last sample, the steps look as follows: 1. 1 2. 2 3. 2 1 4. 3 5. 3 1 6. 3 2 7. 3 2 1 8. 4
instruction
0
5,445
19
10,890
Tags: implementation Correct Solution: ``` n = int(input()) while n != 0: a = 1 k = 0 while a <= n: a *= 2 k += 1 a //= 2 print(k,end=' ') n -= a ```
output
1
5,445
19
10,891
Provide tags and a correct Python 3 solution for this coding contest problem. Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1. You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all already placed slimes. Then, while the last two slimes in the row have the same value v, you combine them together to create a slime with value v + 1. You would like to see what the final state of the row is after you've added all n slimes. Please print the values of the slimes in the row from left to right. Input The first line of the input will contain a single integer, n (1 ≀ n ≀ 100 000). Output Output a single line with k integers, where k is the number of slimes in the row after you've finished the procedure described in the problem statement. The i-th of these numbers should be the value of the i-th slime from the left. Examples Input 1 Output 1 Input 2 Output 2 Input 3 Output 2 1 Input 8 Output 4 Note In the first sample, we only have a single slime with value 1. The final state of the board is just a single slime with value 1. In the second sample, we perform the following steps: Initially we place a single slime in a row by itself. Thus, row is initially 1. Then, we will add another slime. The row is now 1 1. Since two rightmost slimes have the same values, we should replace these slimes with one with value 2. Thus, the final state of the board is 2. In the third sample, after adding the first two slimes, our row is 2. After adding one more slime, the row becomes 2 1. In the last sample, the steps look as follows: 1. 1 2. 2 3. 2 1 4. 3 5. 3 1 6. 3 2 7. 3 2 1 8. 4
instruction
0
5,446
19
10,892
Tags: implementation Correct Solution: ``` n = int(input()) a = [] for i in range(n): a.append(1) while len(a) > 1 and a[len(a) - 1] == a[len(a) - 2]: a.pop() a[len(a) - 1] += 1 for i in a: print(i, end = ' ') ```
output
1
5,446
19
10,893
Provide tags and a correct Python 3 solution for this coding contest problem. Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1. You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all already placed slimes. Then, while the last two slimes in the row have the same value v, you combine them together to create a slime with value v + 1. You would like to see what the final state of the row is after you've added all n slimes. Please print the values of the slimes in the row from left to right. Input The first line of the input will contain a single integer, n (1 ≀ n ≀ 100 000). Output Output a single line with k integers, where k is the number of slimes in the row after you've finished the procedure described in the problem statement. The i-th of these numbers should be the value of the i-th slime from the left. Examples Input 1 Output 1 Input 2 Output 2 Input 3 Output 2 1 Input 8 Output 4 Note In the first sample, we only have a single slime with value 1. The final state of the board is just a single slime with value 1. In the second sample, we perform the following steps: Initially we place a single slime in a row by itself. Thus, row is initially 1. Then, we will add another slime. The row is now 1 1. Since two rightmost slimes have the same values, we should replace these slimes with one with value 2. Thus, the final state of the board is 2. In the third sample, after adding the first two slimes, our row is 2. After adding one more slime, the row becomes 2 1. In the last sample, the steps look as follows: 1. 1 2. 2 3. 2 1 4. 3 5. 3 1 6. 3 2 7. 3 2 1 8. 4
instruction
0
5,447
19
10,894
Tags: implementation Correct Solution: ``` def toBin(num): buf = '' while True: num, rest = num // 2, num - (num // 2 * 2) buf = str(rest)+buf if num == 0: break return int(buf) def main(): num = int(input()) num_bin = toBin(num) num_bin_str = str(num_bin) num_bin_str_len = len(num_bin_str) solutions = [] for pos in range(0, num_bin_str_len): if num_bin_str[pos] == "1": solutions.append(str(num_bin_str_len - pos)) print(" ".join(solutions)) if __name__ == "__main__": main() ```
output
1
5,447
19
10,895
Provide tags and a correct Python 3 solution for this coding contest problem. Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1. You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all already placed slimes. Then, while the last two slimes in the row have the same value v, you combine them together to create a slime with value v + 1. You would like to see what the final state of the row is after you've added all n slimes. Please print the values of the slimes in the row from left to right. Input The first line of the input will contain a single integer, n (1 ≀ n ≀ 100 000). Output Output a single line with k integers, where k is the number of slimes in the row after you've finished the procedure described in the problem statement. The i-th of these numbers should be the value of the i-th slime from the left. Examples Input 1 Output 1 Input 2 Output 2 Input 3 Output 2 1 Input 8 Output 4 Note In the first sample, we only have a single slime with value 1. The final state of the board is just a single slime with value 1. In the second sample, we perform the following steps: Initially we place a single slime in a row by itself. Thus, row is initially 1. Then, we will add another slime. The row is now 1 1. Since two rightmost slimes have the same values, we should replace these slimes with one with value 2. Thus, the final state of the board is 2. In the third sample, after adding the first two slimes, our row is 2. After adding one more slime, the row becomes 2 1. In the last sample, the steps look as follows: 1. 1 2. 2 3. 2 1 4. 3 5. 3 1 6. 3 2 7. 3 2 1 8. 4
instruction
0
5,448
19
10,896
Tags: implementation Correct Solution: ``` n=int(input()) arr=[0]*n i=1 for i in range(n): arr[i]=n%2 n//=2 for j in range(i,-1,-1): if(arr[j]!=0): print(j+1,end=" ") ```
output
1
5,448
19
10,897
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1. You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all already placed slimes. Then, while the last two slimes in the row have the same value v, you combine them together to create a slime with value v + 1. You would like to see what the final state of the row is after you've added all n slimes. Please print the values of the slimes in the row from left to right. Input The first line of the input will contain a single integer, n (1 ≀ n ≀ 100 000). Output Output a single line with k integers, where k is the number of slimes in the row after you've finished the procedure described in the problem statement. The i-th of these numbers should be the value of the i-th slime from the left. Examples Input 1 Output 1 Input 2 Output 2 Input 3 Output 2 1 Input 8 Output 4 Note In the first sample, we only have a single slime with value 1. The final state of the board is just a single slime with value 1. In the second sample, we perform the following steps: Initially we place a single slime in a row by itself. Thus, row is initially 1. Then, we will add another slime. The row is now 1 1. Since two rightmost slimes have the same values, we should replace these slimes with one with value 2. Thus, the final state of the board is 2. In the third sample, after adding the first two slimes, our row is 2. After adding one more slime, the row becomes 2 1. In the last sample, the steps look as follows: 1. 1 2. 2 3. 2 1 4. 3 5. 3 1 6. 3 2 7. 3 2 1 8. 4 Submitted Solution: ``` n=int(input()) b=list(bin(n))[2:] b.reverse() s=str() for i in range(len(b)): if b[i]=="1": s=str(i+1)+" "+s print(s) ```
instruction
0
5,449
19
10,898
Yes
output
1
5,449
19
10,899
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1. You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all already placed slimes. Then, while the last two slimes in the row have the same value v, you combine them together to create a slime with value v + 1. You would like to see what the final state of the row is after you've added all n slimes. Please print the values of the slimes in the row from left to right. Input The first line of the input will contain a single integer, n (1 ≀ n ≀ 100 000). Output Output a single line with k integers, where k is the number of slimes in the row after you've finished the procedure described in the problem statement. The i-th of these numbers should be the value of the i-th slime from the left. Examples Input 1 Output 1 Input 2 Output 2 Input 3 Output 2 1 Input 8 Output 4 Note In the first sample, we only have a single slime with value 1. The final state of the board is just a single slime with value 1. In the second sample, we perform the following steps: Initially we place a single slime in a row by itself. Thus, row is initially 1. Then, we will add another slime. The row is now 1 1. Since two rightmost slimes have the same values, we should replace these slimes with one with value 2. Thus, the final state of the board is 2. In the third sample, after adding the first two slimes, our row is 2. After adding one more slime, the row becomes 2 1. In the last sample, the steps look as follows: 1. 1 2. 2 3. 2 1 4. 3 5. 3 1 6. 3 2 7. 3 2 1 8. 4 Submitted Solution: ``` __author__ = 'Admin' n = int(input()) m = [] for i in range(n): m.append(1) for j in range(i): if m[len(m) - 1] == m[len(m) - 2] and len(m) > 1: m[len(m) - 2] += 1 m.pop(m.index(m[len(m) - 1])) else: break print(*m) ```
instruction
0
5,450
19
10,900
Yes
output
1
5,450
19
10,901
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1. You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all already placed slimes. Then, while the last two slimes in the row have the same value v, you combine them together to create a slime with value v + 1. You would like to see what the final state of the row is after you've added all n slimes. Please print the values of the slimes in the row from left to right. Input The first line of the input will contain a single integer, n (1 ≀ n ≀ 100 000). Output Output a single line with k integers, where k is the number of slimes in the row after you've finished the procedure described in the problem statement. The i-th of these numbers should be the value of the i-th slime from the left. Examples Input 1 Output 1 Input 2 Output 2 Input 3 Output 2 1 Input 8 Output 4 Note In the first sample, we only have a single slime with value 1. The final state of the board is just a single slime with value 1. In the second sample, we perform the following steps: Initially we place a single slime in a row by itself. Thus, row is initially 1. Then, we will add another slime. The row is now 1 1. Since two rightmost slimes have the same values, we should replace these slimes with one with value 2. Thus, the final state of the board is 2. In the third sample, after adding the first two slimes, our row is 2. After adding one more slime, the row becomes 2 1. In the last sample, the steps look as follows: 1. 1 2. 2 3. 2 1 4. 3 5. 3 1 6. 3 2 7. 3 2 1 8. 4 Submitted Solution: ``` n = int(input()); k = 1 a = [] while n > 0 : if n & 1 : a.append(k) k += 1 n = n >> 1 a.reverse() ans = "" for i in range(len(a)) : ans += str(a[i]) + " " print(ans) ```
instruction
0
5,451
19
10,902
Yes
output
1
5,451
19
10,903
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1. You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all already placed slimes. Then, while the last two slimes in the row have the same value v, you combine them together to create a slime with value v + 1. You would like to see what the final state of the row is after you've added all n slimes. Please print the values of the slimes in the row from left to right. Input The first line of the input will contain a single integer, n (1 ≀ n ≀ 100 000). Output Output a single line with k integers, where k is the number of slimes in the row after you've finished the procedure described in the problem statement. The i-th of these numbers should be the value of the i-th slime from the left. Examples Input 1 Output 1 Input 2 Output 2 Input 3 Output 2 1 Input 8 Output 4 Note In the first sample, we only have a single slime with value 1. The final state of the board is just a single slime with value 1. In the second sample, we perform the following steps: Initially we place a single slime in a row by itself. Thus, row is initially 1. Then, we will add another slime. The row is now 1 1. Since two rightmost slimes have the same values, we should replace these slimes with one with value 2. Thus, the final state of the board is 2. In the third sample, after adding the first two slimes, our row is 2. After adding one more slime, the row becomes 2 1. In the last sample, the steps look as follows: 1. 1 2. 2 3. 2 1 4. 3 5. 3 1 6. 3 2 7. 3 2 1 8. 4 Submitted Solution: ``` n=int(input()) d=[] for i in range(n): d.append(1) if len(d)>=2: while(d[-1]==d[-2]): if d[-1]==d[-2]: r=d[-1]+1 d.append(r) d.pop(-2) d.pop(-2) if len(d)<2: break print(*d) ```
instruction
0
5,452
19
10,904
Yes
output
1
5,452
19
10,905
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1. You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all already placed slimes. Then, while the last two slimes in the row have the same value v, you combine them together to create a slime with value v + 1. You would like to see what the final state of the row is after you've added all n slimes. Please print the values of the slimes in the row from left to right. Input The first line of the input will contain a single integer, n (1 ≀ n ≀ 100 000). Output Output a single line with k integers, where k is the number of slimes in the row after you've finished the procedure described in the problem statement. The i-th of these numbers should be the value of the i-th slime from the left. Examples Input 1 Output 1 Input 2 Output 2 Input 3 Output 2 1 Input 8 Output 4 Note In the first sample, we only have a single slime with value 1. The final state of the board is just a single slime with value 1. In the second sample, we perform the following steps: Initially we place a single slime in a row by itself. Thus, row is initially 1. Then, we will add another slime. The row is now 1 1. Since two rightmost slimes have the same values, we should replace these slimes with one with value 2. Thus, the final state of the board is 2. In the third sample, after adding the first two slimes, our row is 2. After adding one more slime, the row becomes 2 1. In the last sample, the steps look as follows: 1. 1 2. 2 3. 2 1 4. 3 5. 3 1 6. 3 2 7. 3 2 1 8. 4 Submitted Solution: ``` f=lambda:map(int,input().split()) n=int(input()) s='1 '*n for i in range(1,n//2): s=s.replace(str(i)+' '+str(i),str(i+1)) print(s) ```
instruction
0
5,453
19
10,906
No
output
1
5,453
19
10,907
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1. You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all already placed slimes. Then, while the last two slimes in the row have the same value v, you combine them together to create a slime with value v + 1. You would like to see what the final state of the row is after you've added all n slimes. Please print the values of the slimes in the row from left to right. Input The first line of the input will contain a single integer, n (1 ≀ n ≀ 100 000). Output Output a single line with k integers, where k is the number of slimes in the row after you've finished the procedure described in the problem statement. The i-th of these numbers should be the value of the i-th slime from the left. Examples Input 1 Output 1 Input 2 Output 2 Input 3 Output 2 1 Input 8 Output 4 Note In the first sample, we only have a single slime with value 1. The final state of the board is just a single slime with value 1. In the second sample, we perform the following steps: Initially we place a single slime in a row by itself. Thus, row is initially 1. Then, we will add another slime. The row is now 1 1. Since two rightmost slimes have the same values, we should replace these slimes with one with value 2. Thus, the final state of the board is 2. In the third sample, after adding the first two slimes, our row is 2. After adding one more slime, the row becomes 2 1. In the last sample, the steps look as follows: 1. 1 2. 2 3. 2 1 4. 3 5. 3 1 6. 3 2 7. 3 2 1 8. 4 Submitted Solution: ``` n = int(input()) s ="1" while n>1: s+="1" n-=1 while len(s)!=len(set(s)): if s[0] == s[1]: s = s.replace(s[0]*2,str(int(s[0])+1)) print(s) ```
instruction
0
5,454
19
10,908
No
output
1
5,454
19
10,909
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1. You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all already placed slimes. Then, while the last two slimes in the row have the same value v, you combine them together to create a slime with value v + 1. You would like to see what the final state of the row is after you've added all n slimes. Please print the values of the slimes in the row from left to right. Input The first line of the input will contain a single integer, n (1 ≀ n ≀ 100 000). Output Output a single line with k integers, where k is the number of slimes in the row after you've finished the procedure described in the problem statement. The i-th of these numbers should be the value of the i-th slime from the left. Examples Input 1 Output 1 Input 2 Output 2 Input 3 Output 2 1 Input 8 Output 4 Note In the first sample, we only have a single slime with value 1. The final state of the board is just a single slime with value 1. In the second sample, we perform the following steps: Initially we place a single slime in a row by itself. Thus, row is initially 1. Then, we will add another slime. The row is now 1 1. Since two rightmost slimes have the same values, we should replace these slimes with one with value 2. Thus, the final state of the board is 2. In the third sample, after adding the first two slimes, our row is 2. After adding one more slime, the row becomes 2 1. In the last sample, the steps look as follows: 1. 1 2. 2 3. 2 1 4. 3 5. 3 1 6. 3 2 7. 3 2 1 8. 4 Submitted Solution: ``` import math n = int(input()) k = math.log2(n) if n % 2 == 0: print(k + 1) else: print(k,' ',1) ```
instruction
0
5,455
19
10,910
No
output
1
5,455
19
10,911
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1. You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all already placed slimes. Then, while the last two slimes in the row have the same value v, you combine them together to create a slime with value v + 1. You would like to see what the final state of the row is after you've added all n slimes. Please print the values of the slimes in the row from left to right. Input The first line of the input will contain a single integer, n (1 ≀ n ≀ 100 000). Output Output a single line with k integers, where k is the number of slimes in the row after you've finished the procedure described in the problem statement. The i-th of these numbers should be the value of the i-th slime from the left. Examples Input 1 Output 1 Input 2 Output 2 Input 3 Output 2 1 Input 8 Output 4 Note In the first sample, we only have a single slime with value 1. The final state of the board is just a single slime with value 1. In the second sample, we perform the following steps: Initially we place a single slime in a row by itself. Thus, row is initially 1. Then, we will add another slime. The row is now 1 1. Since two rightmost slimes have the same values, we should replace these slimes with one with value 2. Thus, the final state of the board is 2. In the third sample, after adding the first two slimes, our row is 2. After adding one more slime, the row becomes 2 1. In the last sample, the steps look as follows: 1. 1 2. 2 3. 2 1 4. 3 5. 3 1 6. 3 2 7. 3 2 1 8. 4 Submitted Solution: ``` import math n = int(input()) log = math.log(n,2) floor = math.floor(log) cnt = 0 arr = [] for i in range(floor,-1,-1): a = 2**i if cnt+a>n: continue else: cnt += a arr.append(a) if cnt==n: break print(*arr) ```
instruction
0
5,456
19
10,912
No
output
1
5,456
19
10,913
Provide a correct Python 3 solution for this coding contest problem. We have a bingo card with a 3\times3 grid. The square at the i-th row from the top and the j-th column from the left contains the number A_{i, j}. The MC will choose N numbers, b_1, b_2, \cdots, b_N. If our bingo sheet contains some of those numbers, we will mark them on our sheet. Determine whether we will have a bingo when the N numbers are chosen, that is, the sheet will contain three marked numbers in a row, column, or diagonal. Constraints * All values in input are integers. * 1 \leq A_{i, j} \leq 100 * A_{i_1, j_1} \neq A_{i_2, j_2} ((i_1, j_1) \neq (i_2, j_2)) * 1 \leq N \leq 10 * 1 \leq b_i \leq 100 * b_i \neq b_j (i \neq j) Input Input is given from Standard Input in the following format: A_{1, 1} A_{1, 2} A_{1, 3} A_{2, 1} A_{2, 2} A_{2, 3} A_{3, 1} A_{3, 2} A_{3, 3} N b_1 \vdots b_N Output If we will have a bingo, print `Yes`; otherwise, print `No`. Examples Input 84 97 66 79 89 11 61 59 7 7 89 7 87 79 24 84 30 Output Yes Input 41 7 46 26 89 2 78 92 8 5 6 45 16 57 17 Output No Input 60 88 34 92 41 43 65 73 48 10 60 43 88 11 48 73 65 41 92 34 Output Yes
instruction
0
5,597
19
11,194
"Correct Solution: ``` A = [[int(x) for x in input().split()] for i in range(3)] n = int(input()) B = set(int(input()) for i in range(n)) res = False for row in A: if all(x in B for x in row): res = True for j in range(3): if all(A[i][j] in B for i in range(3)): res = True if all(A[i][i] in B for i in range(3)): res = True if all(A[i][2 - i] in B for i in range(3)): res = True print("Yes" if res else "No") ```
output
1
5,597
19
11,195
Provide a correct Python 3 solution for this coding contest problem. We have a bingo card with a 3\times3 grid. The square at the i-th row from the top and the j-th column from the left contains the number A_{i, j}. The MC will choose N numbers, b_1, b_2, \cdots, b_N. If our bingo sheet contains some of those numbers, we will mark them on our sheet. Determine whether we will have a bingo when the N numbers are chosen, that is, the sheet will contain three marked numbers in a row, column, or diagonal. Constraints * All values in input are integers. * 1 \leq A_{i, j} \leq 100 * A_{i_1, j_1} \neq A_{i_2, j_2} ((i_1, j_1) \neq (i_2, j_2)) * 1 \leq N \leq 10 * 1 \leq b_i \leq 100 * b_i \neq b_j (i \neq j) Input Input is given from Standard Input in the following format: A_{1, 1} A_{1, 2} A_{1, 3} A_{2, 1} A_{2, 2} A_{2, 3} A_{3, 1} A_{3, 2} A_{3, 3} N b_1 \vdots b_N Output If we will have a bingo, print `Yes`; otherwise, print `No`. Examples Input 84 97 66 79 89 11 61 59 7 7 89 7 87 79 24 84 30 Output Yes Input 41 7 46 26 89 2 78 92 8 5 6 45 16 57 17 Output No Input 60 88 34 92 41 43 65 73 48 10 60 43 88 11 48 73 65 41 92 34 Output Yes
instruction
0
5,598
19
11,196
"Correct Solution: ``` al = [list(map(int, input().split())) for i in range(3)] al = sum(al, []) n = int(input()) bl = [int(input()) for i in range(n)] check = [0 for i in range(9)] for b in bl: if b in al: check[al.index(b)] = 1 if [1,1,1] in [check[:3], check[3:6], check[6:], check[0:7:3], check[1:8:3], check[2:9:3], check[0:9:4], check[2:8:2]]: print("Yes") else: print("No") ```
output
1
5,598
19
11,197
Provide a correct Python 3 solution for this coding contest problem. We have a bingo card with a 3\times3 grid. The square at the i-th row from the top and the j-th column from the left contains the number A_{i, j}. The MC will choose N numbers, b_1, b_2, \cdots, b_N. If our bingo sheet contains some of those numbers, we will mark them on our sheet. Determine whether we will have a bingo when the N numbers are chosen, that is, the sheet will contain three marked numbers in a row, column, or diagonal. Constraints * All values in input are integers. * 1 \leq A_{i, j} \leq 100 * A_{i_1, j_1} \neq A_{i_2, j_2} ((i_1, j_1) \neq (i_2, j_2)) * 1 \leq N \leq 10 * 1 \leq b_i \leq 100 * b_i \neq b_j (i \neq j) Input Input is given from Standard Input in the following format: A_{1, 1} A_{1, 2} A_{1, 3} A_{2, 1} A_{2, 2} A_{2, 3} A_{3, 1} A_{3, 2} A_{3, 3} N b_1 \vdots b_N Output If we will have a bingo, print `Yes`; otherwise, print `No`. Examples Input 84 97 66 79 89 11 61 59 7 7 89 7 87 79 24 84 30 Output Yes Input 41 7 46 26 89 2 78 92 8 5 6 45 16 57 17 Output No Input 60 88 34 92 41 43 65 73 48 10 60 43 88 11 48 73 65 41 92 34 Output Yes
instruction
0
5,599
19
11,198
"Correct Solution: ``` Bingo = [ list(map(int,input().split())) for i in range(3)] n = int(input()) num = [int(input()) for i in range(n)] Bingo.append([Bingo[0][0] , Bingo[1][1],Bingo[2][2] ]) Bingo.append([Bingo[0][2] , Bingo[1][1],Bingo[2][0] ]) for k in range(3): Bingo.append([Bingo[0][k] , Bingo[1][k],Bingo[2][k] ]) k=0 re = 'No' for l in range(len(Bingo)): if len(list(set(num) & set(Bingo[l]))) == 3: re = 'Yes' break print(re) ```
output
1
5,599
19
11,199
Provide a correct Python 3 solution for this coding contest problem. We have a bingo card with a 3\times3 grid. The square at the i-th row from the top and the j-th column from the left contains the number A_{i, j}. The MC will choose N numbers, b_1, b_2, \cdots, b_N. If our bingo sheet contains some of those numbers, we will mark them on our sheet. Determine whether we will have a bingo when the N numbers are chosen, that is, the sheet will contain three marked numbers in a row, column, or diagonal. Constraints * All values in input are integers. * 1 \leq A_{i, j} \leq 100 * A_{i_1, j_1} \neq A_{i_2, j_2} ((i_1, j_1) \neq (i_2, j_2)) * 1 \leq N \leq 10 * 1 \leq b_i \leq 100 * b_i \neq b_j (i \neq j) Input Input is given from Standard Input in the following format: A_{1, 1} A_{1, 2} A_{1, 3} A_{2, 1} A_{2, 2} A_{2, 3} A_{3, 1} A_{3, 2} A_{3, 3} N b_1 \vdots b_N Output If we will have a bingo, print `Yes`; otherwise, print `No`. Examples Input 84 97 66 79 89 11 61 59 7 7 89 7 87 79 24 84 30 Output Yes Input 41 7 46 26 89 2 78 92 8 5 6 45 16 57 17 Output No Input 60 88 34 92 41 43 65 73 48 10 60 43 88 11 48 73 65 41 92 34 Output Yes
instruction
0
5,600
19
11,200
"Correct Solution: ``` A=[0]*9 for i in range(3): A[i*3:i*3+3]=input().split() for i in range(int(input())): b=input() for j in range(9): if A[j]==b: A[j]="0" s="012345678036147258048246" a=f=0 for i in range(24): a+=int(A[int(s[i])]) if i%3==2: if a==0: f+=1 a=0 if f>0: print("Yes") else: print("No") ```
output
1
5,600
19
11,201
Provide a correct Python 3 solution for this coding contest problem. We have a bingo card with a 3\times3 grid. The square at the i-th row from the top and the j-th column from the left contains the number A_{i, j}. The MC will choose N numbers, b_1, b_2, \cdots, b_N. If our bingo sheet contains some of those numbers, we will mark them on our sheet. Determine whether we will have a bingo when the N numbers are chosen, that is, the sheet will contain three marked numbers in a row, column, or diagonal. Constraints * All values in input are integers. * 1 \leq A_{i, j} \leq 100 * A_{i_1, j_1} \neq A_{i_2, j_2} ((i_1, j_1) \neq (i_2, j_2)) * 1 \leq N \leq 10 * 1 \leq b_i \leq 100 * b_i \neq b_j (i \neq j) Input Input is given from Standard Input in the following format: A_{1, 1} A_{1, 2} A_{1, 3} A_{2, 1} A_{2, 2} A_{2, 3} A_{3, 1} A_{3, 2} A_{3, 3} N b_1 \vdots b_N Output If we will have a bingo, print `Yes`; otherwise, print `No`. Examples Input 84 97 66 79 89 11 61 59 7 7 89 7 87 79 24 84 30 Output Yes Input 41 7 46 26 89 2 78 92 8 5 6 45 16 57 17 Output No Input 60 88 34 92 41 43 65 73 48 10 60 43 88 11 48 73 65 41 92 34 Output Yes
instruction
0
5,601
19
11,202
"Correct Solution: ``` A = [] for _ in range(3): for e in list(map(int, input().split())): A.append(e) N = int(input()) for _ in range(N): n = int(input()) if n in A: A[A.index(n)] = 0 patterns = [ [0, 1, 2], [3, 4, 5], [6, 7, 8], [0, 3, 6], [1, 4, 7], [2, 5, 8], [0, 4, 8], [2, 4, 6] ] for pattern in patterns: if not any([A[e] for e in pattern]): print('Yes') exit() print('No') ```
output
1
5,601
19
11,203
Provide a correct Python 3 solution for this coding contest problem. We have a bingo card with a 3\times3 grid. The square at the i-th row from the top and the j-th column from the left contains the number A_{i, j}. The MC will choose N numbers, b_1, b_2, \cdots, b_N. If our bingo sheet contains some of those numbers, we will mark them on our sheet. Determine whether we will have a bingo when the N numbers are chosen, that is, the sheet will contain three marked numbers in a row, column, or diagonal. Constraints * All values in input are integers. * 1 \leq A_{i, j} \leq 100 * A_{i_1, j_1} \neq A_{i_2, j_2} ((i_1, j_1) \neq (i_2, j_2)) * 1 \leq N \leq 10 * 1 \leq b_i \leq 100 * b_i \neq b_j (i \neq j) Input Input is given from Standard Input in the following format: A_{1, 1} A_{1, 2} A_{1, 3} A_{2, 1} A_{2, 2} A_{2, 3} A_{3, 1} A_{3, 2} A_{3, 3} N b_1 \vdots b_N Output If we will have a bingo, print `Yes`; otherwise, print `No`. Examples Input 84 97 66 79 89 11 61 59 7 7 89 7 87 79 24 84 30 Output Yes Input 41 7 46 26 89 2 78 92 8 5 6 45 16 57 17 Output No Input 60 88 34 92 41 43 65 73 48 10 60 43 88 11 48 73 65 41 92 34 Output Yes
instruction
0
5,602
19
11,204
"Correct Solution: ``` a = [] for i in range(3): a += list(map(int, input().split())) bg = [False]*9 n = int(input()) for i in range(n): b = int(input()) if b in a: bg[a.index(b)] = True num = [[0,1,2],[3,4,5],[6,7,8],[0,3,6],[1,4,7],[2,5,8],[0,4,8],[2,4,6]] for x, y, z in num: if bg[x] and bg[y] and bg[z]: print('Yes') exit() print('No') ```
output
1
5,602
19
11,205
Provide a correct Python 3 solution for this coding contest problem. We have a bingo card with a 3\times3 grid. The square at the i-th row from the top and the j-th column from the left contains the number A_{i, j}. The MC will choose N numbers, b_1, b_2, \cdots, b_N. If our bingo sheet contains some of those numbers, we will mark them on our sheet. Determine whether we will have a bingo when the N numbers are chosen, that is, the sheet will contain three marked numbers in a row, column, or diagonal. Constraints * All values in input are integers. * 1 \leq A_{i, j} \leq 100 * A_{i_1, j_1} \neq A_{i_2, j_2} ((i_1, j_1) \neq (i_2, j_2)) * 1 \leq N \leq 10 * 1 \leq b_i \leq 100 * b_i \neq b_j (i \neq j) Input Input is given from Standard Input in the following format: A_{1, 1} A_{1, 2} A_{1, 3} A_{2, 1} A_{2, 2} A_{2, 3} A_{3, 1} A_{3, 2} A_{3, 3} N b_1 \vdots b_N Output If we will have a bingo, print `Yes`; otherwise, print `No`. Examples Input 84 97 66 79 89 11 61 59 7 7 89 7 87 79 24 84 30 Output Yes Input 41 7 46 26 89 2 78 92 8 5 6 45 16 57 17 Output No Input 60 88 34 92 41 43 65 73 48 10 60 43 88 11 48 73 65 41 92 34 Output Yes
instruction
0
5,603
19
11,206
"Correct Solution: ``` B=[] for _ in range(3):B+=list(map(int,input().split())) S=set(int(input()) for _ in range(int(input()))) for i,b in enumerate(B): if b in S:B[i]=0 if B[0]+B[1]+B[2]==0 or B[3]+B[4]+B[5]==0 or B[6]+B[7]+B[8]==0 or B[0]+B[3]+B[6]==0 or B[1]+B[4]+B[7]==0 or B[2]+B[5]+B[8]==0 or B[0]+B[4]+B[8]==0 or B[2]+B[4]+B[6]==0: print('Yes') else:print('No') ```
output
1
5,603
19
11,207
Provide a correct Python 3 solution for this coding contest problem. We have a bingo card with a 3\times3 grid. The square at the i-th row from the top and the j-th column from the left contains the number A_{i, j}. The MC will choose N numbers, b_1, b_2, \cdots, b_N. If our bingo sheet contains some of those numbers, we will mark them on our sheet. Determine whether we will have a bingo when the N numbers are chosen, that is, the sheet will contain three marked numbers in a row, column, or diagonal. Constraints * All values in input are integers. * 1 \leq A_{i, j} \leq 100 * A_{i_1, j_1} \neq A_{i_2, j_2} ((i_1, j_1) \neq (i_2, j_2)) * 1 \leq N \leq 10 * 1 \leq b_i \leq 100 * b_i \neq b_j (i \neq j) Input Input is given from Standard Input in the following format: A_{1, 1} A_{1, 2} A_{1, 3} A_{2, 1} A_{2, 2} A_{2, 3} A_{3, 1} A_{3, 2} A_{3, 3} N b_1 \vdots b_N Output If we will have a bingo, print `Yes`; otherwise, print `No`. Examples Input 84 97 66 79 89 11 61 59 7 7 89 7 87 79 24 84 30 Output Yes Input 41 7 46 26 89 2 78 92 8 5 6 45 16 57 17 Output No Input 60 88 34 92 41 43 65 73 48 10 60 43 88 11 48 73 65 41 92 34 Output Yes
instruction
0
5,604
19
11,208
"Correct Solution: ``` *inputs, = map(int, open(0).read().split()) A = inputs[:9] B = inputs[9:] C = [0] * 9 for b in B: if b in A: C[A.index(b)] = 1 if any([ all(C[:3]), all(C[3:6]), all(C[6:]), all(C[::3]), all(C[1::3]), all(C[2::3]), all([C[0], C[4], C[8]]), all([C[2], C[4], C[6]]) ]): print('Yes') else: print('No') ```
output
1
5,604
19
11,209
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. We have a bingo card with a 3\times3 grid. The square at the i-th row from the top and the j-th column from the left contains the number A_{i, j}. The MC will choose N numbers, b_1, b_2, \cdots, b_N. If our bingo sheet contains some of those numbers, we will mark them on our sheet. Determine whether we will have a bingo when the N numbers are chosen, that is, the sheet will contain three marked numbers in a row, column, or diagonal. Constraints * All values in input are integers. * 1 \leq A_{i, j} \leq 100 * A_{i_1, j_1} \neq A_{i_2, j_2} ((i_1, j_1) \neq (i_2, j_2)) * 1 \leq N \leq 10 * 1 \leq b_i \leq 100 * b_i \neq b_j (i \neq j) Input Input is given from Standard Input in the following format: A_{1, 1} A_{1, 2} A_{1, 3} A_{2, 1} A_{2, 2} A_{2, 3} A_{3, 1} A_{3, 2} A_{3, 3} N b_1 \vdots b_N Output If we will have a bingo, print `Yes`; otherwise, print `No`. Examples Input 84 97 66 79 89 11 61 59 7 7 89 7 87 79 24 84 30 Output Yes Input 41 7 46 26 89 2 78 92 8 5 6 45 16 57 17 Output No Input 60 88 34 92 41 43 65 73 48 10 60 43 88 11 48 73 65 41 92 34 Output Yes Submitted Solution: ``` A=[] for i in [0]*3: A+=list(map(int,input().split())) N=int(input()) B=[0]*9 for i in range(N): i=int(input()) if i in A: B[A.index(i)]=1 res=0 for i in range(3): res+=B[3*i]*B[3*i+1]*B[3*i+2] res+=B[i]*B[i+3]*B[i+6] res+=B[0]*B[4]*B[8]+B[2]*B[4]*B[6] print(['No','Yes'][res>0]) ```
instruction
0
5,605
19
11,210
Yes
output
1
5,605
19
11,211
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. We have a bingo card with a 3\times3 grid. The square at the i-th row from the top and the j-th column from the left contains the number A_{i, j}. The MC will choose N numbers, b_1, b_2, \cdots, b_N. If our bingo sheet contains some of those numbers, we will mark them on our sheet. Determine whether we will have a bingo when the N numbers are chosen, that is, the sheet will contain three marked numbers in a row, column, or diagonal. Constraints * All values in input are integers. * 1 \leq A_{i, j} \leq 100 * A_{i_1, j_1} \neq A_{i_2, j_2} ((i_1, j_1) \neq (i_2, j_2)) * 1 \leq N \leq 10 * 1 \leq b_i \leq 100 * b_i \neq b_j (i \neq j) Input Input is given from Standard Input in the following format: A_{1, 1} A_{1, 2} A_{1, 3} A_{2, 1} A_{2, 2} A_{2, 3} A_{3, 1} A_{3, 2} A_{3, 3} N b_1 \vdots b_N Output If we will have a bingo, print `Yes`; otherwise, print `No`. Examples Input 84 97 66 79 89 11 61 59 7 7 89 7 87 79 24 84 30 Output Yes Input 41 7 46 26 89 2 78 92 8 5 6 45 16 57 17 Output No Input 60 88 34 92 41 43 65 73 48 10 60 43 88 11 48 73 65 41 92 34 Output Yes Submitted Solution: ``` f=lambda a:any(all(b)for b in a)|all(a[i][i]for i in(0,1,2)) *t,=map(int,open(0).read().split()) a=t[:9] s=eval('[0]*3,'*3) for b in t[10:]: if b in a: i=a.index(b) s[i//3][i%3]=1 print('NYoe s'[f(s)|f([t[::-1]for t in zip(*s)])::2]) ```
instruction
0
5,606
19
11,212
Yes
output
1
5,606
19
11,213
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. We have a bingo card with a 3\times3 grid. The square at the i-th row from the top and the j-th column from the left contains the number A_{i, j}. The MC will choose N numbers, b_1, b_2, \cdots, b_N. If our bingo sheet contains some of those numbers, we will mark them on our sheet. Determine whether we will have a bingo when the N numbers are chosen, that is, the sheet will contain three marked numbers in a row, column, or diagonal. Constraints * All values in input are integers. * 1 \leq A_{i, j} \leq 100 * A_{i_1, j_1} \neq A_{i_2, j_2} ((i_1, j_1) \neq (i_2, j_2)) * 1 \leq N \leq 10 * 1 \leq b_i \leq 100 * b_i \neq b_j (i \neq j) Input Input is given from Standard Input in the following format: A_{1, 1} A_{1, 2} A_{1, 3} A_{2, 1} A_{2, 2} A_{2, 3} A_{3, 1} A_{3, 2} A_{3, 3} N b_1 \vdots b_N Output If we will have a bingo, print `Yes`; otherwise, print `No`. Examples Input 84 97 66 79 89 11 61 59 7 7 89 7 87 79 24 84 30 Output Yes Input 41 7 46 26 89 2 78 92 8 5 6 45 16 57 17 Output No Input 60 88 34 92 41 43 65 73 48 10 60 43 88 11 48 73 65 41 92 34 Output Yes Submitted Solution: ``` import sys A=[] for a in range(3): l=[int(i) for i in input().split()] A.extend(l) f=[0]*9 for i in range(int(input())): b=int(input()) if b in A: f[A.index(b)]=1 #print(f) p=[[0,1,2],[3,4,5],[6,7,8],[0,3,6],[1,4,7],[2,5,8],[0,4,8],[2,4,6]] for i in p: if all(f[x]==1 for x in i): print("Yes") sys.exit() print("No") ```
instruction
0
5,607
19
11,214
Yes
output
1
5,607
19
11,215
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. We have a bingo card with a 3\times3 grid. The square at the i-th row from the top and the j-th column from the left contains the number A_{i, j}. The MC will choose N numbers, b_1, b_2, \cdots, b_N. If our bingo sheet contains some of those numbers, we will mark them on our sheet. Determine whether we will have a bingo when the N numbers are chosen, that is, the sheet will contain three marked numbers in a row, column, or diagonal. Constraints * All values in input are integers. * 1 \leq A_{i, j} \leq 100 * A_{i_1, j_1} \neq A_{i_2, j_2} ((i_1, j_1) \neq (i_2, j_2)) * 1 \leq N \leq 10 * 1 \leq b_i \leq 100 * b_i \neq b_j (i \neq j) Input Input is given from Standard Input in the following format: A_{1, 1} A_{1, 2} A_{1, 3} A_{2, 1} A_{2, 2} A_{2, 3} A_{3, 1} A_{3, 2} A_{3, 3} N b_1 \vdots b_N Output If we will have a bingo, print `Yes`; otherwise, print `No`. Examples Input 84 97 66 79 89 11 61 59 7 7 89 7 87 79 24 84 30 Output Yes Input 41 7 46 26 89 2 78 92 8 5 6 45 16 57 17 Output No Input 60 88 34 92 41 43 65 73 48 10 60 43 88 11 48 73 65 41 92 34 Output Yes Submitted Solution: ``` a=[list(map(int,input().split())) for _ in range(3)] f=[[0]*3 for _ in range(3)] for _ in range(int(input())): x=int(input()) for i in range(3): for j in range(3): if x==a[i][j]: f[i][j]=1 ans=0 for i in range(3): if all(f[i][j] for j in range(3)) or all(f[j][i] for j in range(3)): ans=1 if f[0][0]==f[1][1]==f[2][2]==1 or f[0][2]==f[1][1]==f[2][0]==1: ans=1 print("Yes" if ans else "No") ```
instruction
0
5,608
19
11,216
Yes
output
1
5,608
19
11,217
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. We have a bingo card with a 3\times3 grid. The square at the i-th row from the top and the j-th column from the left contains the number A_{i, j}. The MC will choose N numbers, b_1, b_2, \cdots, b_N. If our bingo sheet contains some of those numbers, we will mark them on our sheet. Determine whether we will have a bingo when the N numbers are chosen, that is, the sheet will contain three marked numbers in a row, column, or diagonal. Constraints * All values in input are integers. * 1 \leq A_{i, j} \leq 100 * A_{i_1, j_1} \neq A_{i_2, j_2} ((i_1, j_1) \neq (i_2, j_2)) * 1 \leq N \leq 10 * 1 \leq b_i \leq 100 * b_i \neq b_j (i \neq j) Input Input is given from Standard Input in the following format: A_{1, 1} A_{1, 2} A_{1, 3} A_{2, 1} A_{2, 2} A_{2, 3} A_{3, 1} A_{3, 2} A_{3, 3} N b_1 \vdots b_N Output If we will have a bingo, print `Yes`; otherwise, print `No`. Examples Input 84 97 66 79 89 11 61 59 7 7 89 7 87 79 24 84 30 Output Yes Input 41 7 46 26 89 2 78 92 8 5 6 45 16 57 17 Output No Input 60 88 34 92 41 43 65 73 48 10 60 43 88 11 48 73 65 41 92 34 Output Yes Submitted Solution: ``` A = [[0,0,0],[0,0,0],[0,0,0]] for i in range(3): A[i][0],A[i][1],A[i][2] = map(int, input().split()) N = int(input()) L = list() Al = [i for x in A for i in x] have = [] for _ in range(N): a = int(input()) if a in Al: have.append(a) if len(have)>=4: print("Yes") else: print("No") ```
instruction
0
5,609
19
11,218
No
output
1
5,609
19
11,219